What is the perimeter of ABCD?

Answer:

Step-by-step explanation:

c(25) = 85(25) + 600 = 2725

Answer:

What’s the answer?!? Ok

Step-by-step explanation:

Answer:

ok.

1. Tanya spends exactly $30.

She can spend $15.00 each for 2 t shirts and spend $30.00

15 x 2 = 30

She can buy one $15.00 t shirt, one $10.00 t shirt and one $5.00 t shirt for exactly $30.00.

15 + 10 + 5 = 30

She can buy one $15.00 t shirt, and three $5.00 t shirts for exactly $30.00

15 + 5 + 5 +5 = 30

2. To spend exactly $25.00, Tanya can spend it :

5 of the $5.00 t shirts

5 x 5 = 25

2 of the $10 t shirts and one of the $5 t shirts

10 + 10 + 5 = 25

1 of the $15 t shirts and one of the $10 t shirts

15 + 10 = 25

1 of the $15 t shirts and 2 of the $5 t shirts

15 + 5 + 5 = 25

Answer:

3+5+4=12units3

Step-by-step explanation:

A prism has a length of 3 units, height of 5 units, and width of 4 units. 3 + 5 + 4 = 12 Units cubed 3(4) + 5 = 17 Units cubed.

Answer:

He is wrong its really 60 units

Step-by-step explanation:

Answer:

a.)Q' = -kQ + r

b.)Q = [tex]\frac{r}{k} [ 1 - e^{-kt} ][/tex]

c.) Limiting long run value of Q = [tex]\frac{r}{k}[/tex]

Step-by-step explanation:

a.)

The rate of change is directly proportional to the quantity, Q

⇒[tex]\frac{dQ}{dt}[/tex] ∝ Q

⇒[tex]\frac{dQ}{dt}[/tex] = -kQ ( because it is decreasing )

Also given, quantity is increasing with a constant rate r

⇒[tex]\frac{dQ}{dt}[/tex] = -kQ + r

⇒Q' = -kQ + r

b.)

As we have

[tex]\frac{dQ}{dt}[/tex] = -kQ + r

⇒[tex]\frac{dQ}{-kQ + r}[/tex] = dt

⇒∫[tex]\frac{dQ}{-kQ + r}[/tex] = ∫dt

⇒-[tex]\frac{1}{k}log(-kQ + r) = t + C[/tex]

⇒log(-kQ + r) = -kt -kC

Taking exponential both side, we get

⇒-kQ + r = [tex]e^{-kt +A}[/tex]

⇒-kQ = -r + [tex]e^{-kt +A}[/tex]

⇒Q = [tex]\frac{r}{k}[/tex] - [tex]\frac{1}{k}[/tex][tex]e^{-kt +A}[/tex]

⇒Q = [tex]\frac{r}{k}[/tex] - [tex]\frac{1}{k}[/tex][tex]e^{-kt}.e^{A}[/tex]

⇒Q = [tex]\frac{r}{k}[/tex] - [tex]\frac{1}{k}[/tex][tex]Be^{-kt}[/tex]      .......(1)

Now,

At t = 0, Q = 0

0 = [tex]\frac{r}{k}[/tex] - [tex]\frac{1}{k}[/tex]B

⇒[tex]\frac{1}{k}B = \frac{r}{k}[/tex]

⇒B = r

∴ equation (1) becomes

Q = [tex]\frac{r}{k}[/tex] - [tex]\frac{r}{k}[/tex][tex]e^{-kt}[/tex]

⇒Q = [tex]\frac{r}{k} [ 1 - e^{-kt} ][/tex]

c.)

for limiting long run value of Q

[tex]\lim_{n \to \infty} Q = \lim_{n \to \infty}[/tex]   [tex]\frac{r}{k} [ 1 - e^{-kt} ][/tex]

               =  [tex]\frac{r}{k}[/tex][tex]\lim_{n \to \infty} [ 1 - e^{-kt} ]= \frac{r}{k} [ 1 - e^{\infty} ] = \frac{r}{k}[ 1-0][/tex]

               = [tex]\frac{r}{k}[/tex]

⇒Limiting long run value of Q = [tex]\frac{r}{k}[/tex]

Q' = -kQ + r

Q = r/k (1- [tex]\rm e^{-kt}[/tex])

The limiting long-run value of Q = [tex]\rm \frac{r}{k}[/tex]

A drug is administered intravenously at a constant rate of r mg/hour and is excreted at a rate proportional to the quantity present, with a constant of proportionality k > 0.

The differential equation is an equation that contains the derivative of the unknown function.

a) It is given that the rate of change is directly proportional to the quantity Q

[tex]\rm \frac{dQ}{dt}[/tex] ∝ Q

[tex]\rm \frac{dQ}{dt}[/tex] = -kQ

So, Q' = -kQ + r

where quantity Q is increasing with constant rate r and k is unknown constant.

b) Q' = -kQ + r

[tex]\rm \frac{dQ}{dt}[/tex] = -kQ + r

[tex]\rm \frac{dQ}{-kQ+r}=dt\\[/tex]

[tex]\int\limits {\rm \frac{dQ}{-kQ+r} = \int\limits dt[/tex]

log(-kQ + r) = -kt -kC

by taking exponential both side, we get

-kQ + r = [tex]e^{-kt+A}[/tex]

-kQ = -r + [tex]e^{-kt+A}[/tex]

-Q /k= -r/k + [tex]e^{-kt+A}[/tex]/k

Q = -r/k -[tex]\rm \frac{1}{k}Be^{-kt}[/tex]

At t = 0, Q = 0

[tex]\rm \frac{1}{k} B=\frac{r}{k}[/tex]

B = r

by substituting the value in the above equation

Q = r/k (1- [tex]\rm e^{-kt}[/tex])

c) The limiting long-run value of Q

[tex]\rm \lim_{n \to \infty} Q= \lim_{n \to \infty} \frac{r}{k} [1-e^{-kt} ]\\\rm= \frac{r}{k}\lim_{n \to \infty} [1-e^{-kt} ]\\\rm =\frac{r}{k}[1-0]\\\rm=\frac{r}{k}[/tex]

The value Q =[tex]\rm \frac{r}{k}[/tex]

Learn more about differential equations here:

https://brainly.com/question/25731911

Answer:

The recurring decimal 0.004 in friction is 1/250

Answer:

Step-by-step explanation:

Let x=0.004444....

1000x-100x=4.444... -0.444...

900x=4

x=4/900=1/225

Answer:

A

Step-by-step explanation:

The fact you need to use is 47%

47/100 * 47/100 = 2209 / 10000 = 0.221

Answer:

50

Step-by-step explanation:

25/100 × 200

= 25×2

=50

Answer:

What impact did the American Revolution have on Haiti? The American Revolution occurred first and in part inspired the Haitian and French Revolutions. ... The Haitian Revolution was the first and only slave uprising that led to the establishment of a free state without slavery and ruled by non-whites and former slaves.

What impact did the Haitian Revolution have on the United States?The Haitian Revolution had many international repercussions. It ended Napoleon's attempts to create a French empire in the Western Hemisphere and arguably caused France to decide to sell its North American holdings to the United States (the Louisiana Purchase)—thus enabling the expansion of slavery into that territory.

Step-by-step explanation:

BRAINLIST PLEASE!!

Answer:

acute angle

Step-by-step explanation:

Big brain

Answer:

Adjacent

Step-by-step explanation:

Answer:

15000(1.003425)^12t ;

4.11%

4.188%

Step-by-step explanation:

Given that:

Loan amount = principal = $15000

Interest rate, r = 4.11% = 0.0411

n = number of times compounded per period, monthly = 12 (number of months in a year)

Total amount, F owed, after t years in college ;

F(t) = P(1 + r/n)^nt

F(t) = 15000(1 + 0.0411/12)^12t

F(t) = 15000(1.003425)^12t

2.) The annual percentage rate is the interest rate without compounding = 4.11%

3.)

The APY

APY = (1 + APR/n)^n - 1

APY = (1 + 0.0411/12)^12 - 1

APY = (1.003425)^12 - 1

APY = 1.04188 - 1

APY = 0.04188

APY = 0.04188 * 100% = 4.188%

Answer:

✔️The graph contains point zero

✔️There's a closed circle on 2⁸/9

✔️The arrow points to the left

Step-by-step explanation:

The inequality, 2⁸/9 ≥ x, implies that 2⁸/9 is either equal to x or more than x. In other words, the highest value of x is 2⁸/9. Possible value of x is equal to or less than 2⁸/9.

This, on a number line, we would graph the information as follows:

✔️The graph contains point zero

Rationale: point zero is less than 2⁸/9

✔️There's a closed circle on 2⁸/9

Rationale: this indicates 2⁸/9 is included in the possible value of x

✔️The arrow points to the left

Rationale: possible value of x ranges from 2⁸/9 and below.

Answer: b c d

Step-by-step explanation:

heart it please ;) your welcome

Answer:

what is your question

Step-by-step explanation: am not trying to steal ur points

Answer: 3.9

Step-by-step explanation:

7.8 miles x 0.5 = 3.9 miles

Answer:

x = 10 since 10 x 10 = 100

Step-by-step explanation:

Answer:2

Step-by-step explanation:

Answer:

12

Step-by-step explanation:

its either 12 or 18 try 12 first

Answer:

8 times,

Step-by-step explanation:

4 x 8 is 32

Answer:

El raigañu cuadráu corresponder cola radicación d'índiz 2 o, equivalentemente, cola potenciación d'esponente 1/2.

Cualquier númberu real non negativu {\displaystyle x}x tien un únicu raigañu cuadráu positiva o raigañu cuadráu principal[2] y denotada como {\displaystyle {\sqrt {x}}}{\displaystyle {\sqrt {x}}} onde {\displaystyle {\sqrt {\;}}}{\displaystyle {\sqrt {\;}}} ye'l símbolu raigañu y {\displaystyle x}x ye'l aniciando.

Cuando se riquir denotar dos raigaños cuadraos una negativa, {\displaystyle -{\sqrt {x}}}{\displaystyle -{\sqrt {x}}}, y otra positiva, {\displaystyle {\sqrt {x}}}{\displaystyle {\sqrt {x}}}, suelen denotarse curioso como {\displaystyle \pm {\sqrt {x}}}{\displaystyle \pm {\sqrt {x}}} o bien como {\displaystyle \mp {\sqrt {x}}}{\displaystyle \mp {\sqrt {x}}} según l'orde precisáu.

El conceutu de raigañu cuadráu puede estendese a cualquier aniellu alxebraicu, asina ye posible definir el raigañu cuadráu d'un númberu real negativu o'l raigañu cuadráu de delles matrices. Nos númberos cuaterniónicos los reales negativos almiten un númberu infinitu de raigaños cuadraos, sicasí'l restu de cuaterniones distintos de cero almiten solu dos raigaños cuadraos. Nel aniellu non conmutativu de les funciones reales de variable real cola adición y la composición de funciones si fºf = g, puede plantegase que f ye la "raigañu cuadráu" de g.[3]

Step-by-step explanationY segun todos mis calculos es 2 :

Answer:

2 hours 52 minutes

Step-by-step explanation:

11-2=9

26/9=2.87

Answer:

primero sumas todas las calificaciones

posteriormente las divedes entre las materias que son y listo sacas su promedio

espero y te ayude

Answer:

1.15m/s

Step-by-step explanation:

Given data

m1=0.025 kg

m2= 2.35kg

u1= 217 m/s

v1= 109 m/s

u2= 0

v2=???

Applying the law of conservation of linear momentum

m1u1+m2u2=m1v1+m2v2

substitute

0.025*217+2.35*0= 0.025*109+2.35*v2

5.425= 2.725+ 2.35v2

5.425-2.725=2.35v2

2.7=2.35v2

divide both sides by 2.35

v2= 2.7/2.35

v2=1.15 m/s

Hence the speed of the watermelon will be 1.15m/s

Answer:

Step-by-step explanation:

12 + 4b = 72

Subtract 12 on both sides

4b = 60

b = 15

Answer:

C. 18x+9y

Step-by-step explanation:

Answer:

I think its the third one, C. Hope this helps!

Answer:

1. HL

2. ASA

3. AAS

4. SSS

Step-by-step explanation:

1. The hypotenuse and right angle of one triangle is congruent to the other triangle's hypotenuse and right angle, proving the hypotenuse-leg congruency theorem (HL).

2. The two angles and their included side (the side between the two angles) of a triangle match those of the other triangle, proving the angle-side-angle congruency theorem (ASA).

3. The measures of the side and the two adjacent angles of a triangle are given. They are congruent to the corresponding parts of the other triangle, proving the angle-angle-side congruency theorem (AAS).

4. All side lengths of one triangle are given, matching the corresponding sides of the other triangle. This proves the side-side-side congruency theorem (SSS).

I hope this helped!

Answer:

The mean is 95 and the standard deviation is 2

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question:

Population:

Mean 95, Standard deviation 12

Samples of size 36:

By the Central Limit Theorem,

Mean 95

Standard deviation [tex]s = \frac{12}{\sqrt{36}} = 2[/tex]

Answer:

The relationship between d, c and p is that [tex]p = \frac{4c^2}{9d}[/tex]

Step-by-step explanation:

Roots of a quadratic equation:

The roots [tex]r_1[/tex] and [tex]r_2[/tex] of a quadratic equation in the following format:

[tex]ax^2 + bx + c = 0[/tex]

Can be given by:

[tex]r_1 + r_2 = -\frac{b}{a}[/tex]

[tex]r_1r_2 = \frac{c}{a}[/tex]

In this question:

We have the following quadratic equation:

[tex]dx^2 - cx + p = 0[/tex]

So [tex]a = d, b = -c, c = p[/tex]

One of the roots is twice the other:

So [tex]r_2 = 2r_1[/tex]

First relation:

[tex]r_1 + r_2 = -\frac{b}{a}[/tex]

[tex]r_1 + 2r_1 = -\frac{(-c)}{d}[/tex]

[tex]3r_1 = \frac{c}{d}[/tex]

[tex]r_1 = \frac{c}{3d}[/tex]

Second relation:

[tex]r_1r_2 = \frac{c}{a}[/tex]

[tex]r_1*2r_1 = \frac{p}{d}[/tex]

[tex]2r_{1}^{2} = \frac{p}{d}[/tex]

[tex](\frac{2c}{3d})^2 = \frac{p}{d}[/tex]

[tex]\frac{4c^2}{9d^2} = \frac{p}{d}[/tex]

[tex]p = \frac{4c^2}{9d}[/tex]

The relationship between d, c and p is that [tex]p = \frac{4c^2}{9d}[/tex]

Answer:

a. (5,5)

b. (-5,-5)

Step-by-step explanation:

reflection over x-axis flips the sign of y-coordinate

reflection over y-axis flips the sign of x-coordinate