the measures of the angles of a triangle are shown in the figure. solve for x

Answer:

[tex]A=-323[/tex]

Step-by-step explanation:

A = (-123)+77+(-257)+23-43

A = -46+(-257)+23-43

A = -303+23-43

A = -280-43

A = -323

Hope this helps

Answer:

9x^2 + 27x + 18

Step-by-step explanation:

(9x + 9)(x + 2)

=9x(x + 2) +9(x + 2)

=9x^2 + 18x + 9x + 18

=9x^2 + 27x + 18

Answer:

Step-by-step explanation:

Answer:

∠RPQ = 27

Step-by-step explanation:

In ΔSRQ,

∠R = 90

∠SQR = 36°

∠R + ∠SQR + ∠RSQ = 180    {Angle sum property of triangle}

90 + 36 + ∠RSQ = 180

126 + ∠RSQ = 180

         ∠RSQ = 180 - 126

        ∠RSQ = 54°

∠PSQ +∠RSQ = 180   {Linear pair}

∠PSQ + 54 = 180

 ∠PSQ = 180 - 54

∠PSQ = 126

In ΔPSQ,

SQ = PS ,

So, ∠SQP = ∠SPQ    {Angles opposite to equal sides are equal}

∠SQP = ∠SPQ   =x

∠PSQ + x +x = 180  {Angle sum property of triangle}

126 + 2x = 180

2x = 180 - 126

2x = 54

x = 54/2

x = 27

∠RPQ = 27°

Answer:

[tex]x^2+4x-7=(x+4)^2-23[/tex] where [tex]a=4[/tex] and [tex]b=-23[/tex]

Step-by-step explanation:

Given: [tex]x^2+4x-7[/tex]

Complete the square: [tex](x^2+4x+16)-23[/tex]

Reduce trinomial: [tex](x+4)^2-23[/tex]

Therefore, [tex]x^2+4x-7=(x+4)^2-23[/tex] where [tex]a=4[/tex] and [tex]b=-23[/tex].

Answer:

Instrumental variable

Step-by-step explanation:

cov(z,u)  shows that the instrumental variable is exogenous. An instrumental variable can also be regarded as a third variable, It is included in the model if you have an issue of endogeneity in your regression model. An endogeneity problem arises if there are other variables are being influenced by other independent variable in the same regression model. we use z to account for these behaviours in the model. this variable controls for measurement errors as well as confounding errors.  

Answer:

see below

Step-by-step explanation:

-X – 10 > 14 + 2x.

Subtract 2x from each side

-x-2x -10 > 14+2x-2x

-3x -10 > 14

Add 10 to each side

-3x -10+10 > 14+10

-3x > 24

Divide each side by -3, remembering to flip the inequality

-3x/-3 < 24/-3

x < -8

Answer:

x = 2

Step-by-step explanation:

Given

2.5(6x - 4) = 10 + 4(1.5 + 0.5x) ← distribute parenthesis on both sides

15x - 10 = 10 + 6 + 2x , that is

15x - 10 = 16 + 2x ( subtract 2x from both sides )

13x - 10 = 16 ( add 10 to both sides )

13x = 26 ( divide both sides by 13 )

x = 2

Answer:

2

Step-by-step explanation:

15x-10=10+6+2x

15x-2x=10+10+6

13x=26

divide 13 both sides

x=2

We are given the linear function below:

[tex] \large{f(x) = 3x + 13}[/tex]

To find f(4), simply substitute x = 4 in the equation.

[tex] \large{f(4) = 3(4) + 13} \\ \large{f(4) = 12 + 13} \\ \large \boxed{f(4) = 25}[/tex]

Answer

Answer:

x + 5 = 3x -1

7x + 21 = 5x + 35

5x + 5 = 6x - 5

Step-by-step explanation:

7x+21=5x+35

-5x -5x

2x+21=35

-21 -21

2x=14

÷2 ÷2

x=7

x+5=3x-1

-x -x

5=2x-1

+1 +1

6=2x

÷2 ÷2

x=3

5x+5=6x-5

-5x -5x

5=x-5

+5 +5

x=10

Hope this helps! :)

Answer:

(3, 3) (1, -1)

Step-by-step explanation:

Answer:

(1, - 1) , (3, 3 )

Step-by-step explanation:

Given the 2 equations

3 + y - 2x = 0 → (1)

3x² + 2y² - 4xy = 9 → (2)

In (1) subtract 3 - 2x from both sides

y = 2x - 3 → (3)

Substitute y = 2x - 3 into (2)

3x² + 2(2x - 3)² - 4x(2x - 3) = 9 ← distribute and simplify left side

3x² + 2(4x² - 12x + 9) - 8x² + 12x = 9

3x² + 8x² - 24x + 18 - 8x² + 12x = 0

3x² - 12x + 18 = 9 ( subtract 9 from both sides )

3x² - 12x + 9 = 0 ( divide through by 3 )

x² - 4x + 3 = 0 ← in standard form

(x - 1)(x - 3) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 1 = 0 ⇒ x = 1

x - 3 = 0 ⇒ x = 3

Substitute these values into (3) for corresponding values of y

x = 1 : y = 2(1) - 3 = 2 - 3 = - 1 ⇒ (1, - 1 )

x = 3 : y = 2(3) - 3 = 6 - 3 = 3 ⇒ (3, 3 )

Answer:

Step-by-step explanation:

Answer:

2y

Step-by-step explanation:

2x-3=2y

2x-2y=3

Answer:

x = 3, y = 1.5

Step-by-step explanation:

Given the 2 equations

x + 2y = 6 → (1)

2x - 3 = 2y → (2)

In (2) subtract 2y from both sides and add 3 to both sides

2x - 2y = 3 → (3)

Adding (1) and (3) term by term will eliminate the y- term

3x = 9 ( divide both sides by 3 )

x = 3

Substitute x = 3 into either of the 2 equations and solve for y

Substituting into (1)

3 + 2y = 6 ( subtract 3 from both sides )

2y = 3 ( divide both sides by 2 )

y = 1.5

Answer:

i think its A but im not sure its correct

Step-by-step explanation:

Answer:

Mean A=7

median B=3/4 .75

Answer:

Step-by-step explanation:

mean = sum of data / no of data

= 4+7+5+9+11+6 /6

=42 / 6

=7

mean = 7

to find median arrange the data from ascending to descending order

Data = 3,4,6,8,10,12

median = (N +1) /      (N means no of data)

=(6 +1) / 2

=7/2

=3.5

= 3 rd term + 4 th term / 2

=6 + 8 / 2

=14 / 2

=7

median = 7

Answer:

18 i think sorry if i am wrong

Step-by-step explanation:

Answer:

s

Step-by-step explanation:

Answer:

[tex]\frac{3}{4}[/tex]

Step-by-step explanation:

Vowels: A, E, I, O, U

There are only 2 vowels out of 8 on the spinner, which means that 6 out of 8 aren't vowels.

[tex]\frac{6}{8}[/tex] = [tex]\frac{3}{4}[/tex]

Hope this is helpful

Answer:

12/37

Step-by-step explanation:

SOH - CAH - TOA

O = opposite

A = adjacent

H = hypotenuse

Sin = O/H

Sin N = 12/37

Answer:Find sin N

sinN=

hypotenuse/opposite

​12/37

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

Answer:

Step-by-step explanation:

             Statements                                  Reasons

1). AD ≅ BC                                           1). Given

2). ∠D ≅ ∠C                                          2). Given

3). ∠DEA ≅ ∠CEB                                 3). Vertically opposite angles

4). ΔADE ≅ Δ BCE                                4). AAS postulate of congruence

5). AE ≅ BE                                           5). CPCTC

Hence, proved.

We have a linear function y=kx+b and we know that                                                       [tex]\displaystyle\bf \left \{ {{5*1+c=2} \atop {5p+c=-2}} \right.=>\left \{ {{c=-3} \atop {5p+c=-2}} \right. => 5p-3=-2 => p=\frac{1}{5}[/tex]                                       [tex]\displaystyle\bf Answer: \boxed{C) \quad p=\frac{1}{5} }[/tex]                            

Answer:

maybe y - -3x = -11?

Step-by-step explanation:

im rushing so i may have did it wrong sorry

Answer:

Assuming you want to know there was for UK it was [tex]\frac{3}{8}[/tex]

Step-by-step explanation:

Let's add the fractions together

[tex]\frac{1}{3} +\frac{1}{6}+\frac{1}{8\\}\\\frac{8}{24} +\frac{4}{24} +\frac{3}{24} =\frac{15}{24}[/tex]

The remainder is the amount in UK so

[tex]\frac{24}{24} -\frac{15}{24} =\frac{9}{24} \\[/tex]

Simplified it's [tex]\frac{3}{8}[/tex]

9514 1404 393

Answer:

  x = -1/3

Step-by-step explanation:

Maybe you mean ...

  (1/49)^(x +2) = 7^(x -3)

Taking logarithms base 7, we have ...

  -2(x +2) = x -3

  -2x -4 = x -3 . . . . . eliminate parentheses

  -1 = 3x . . . . . . . . . . add 3+2x

  x = -1/3 . . . . . . . . . divide by the coefficient of x

This is confirmed by a calculator in the first attachment.

_____

As written, the equation is ...

  ((1/49)^x) +2 = (7^x) -3

Multiplying both sides by 7^(2x) gives us ...

  1 + 2·7^(2x) = 7^(3x) -3·7^(2x)

Subtracting the left side gives us the cubic equation in 7^x:

  (7^x)^3 -5·(7^x)^2  -1 = 0

This has one irrational real root (found using a graphing calculator). The solution is ...

  x ≈ 0.83111881051

This answer is confirmed in the second attachment. (10^-16 is essentially zero for this purpose.)

__

Additional comment

Part of the reason for going to the trouble to solve the "as written" equation is to show you that it is possible, though messy. What we'd like to emphasize here is that grouping symbols are needed when exponents contain math operations.

In typeset text, the superscript font indicates the grouping of parts of an exponent. In plain text, we cannot tell if 7^x-3 means (7^x)-3 or 7^(x-3). The Order of Operations tells us to interpret it the first way: (7^x) -3, because exponentiation is done before addition. That leads to trouble in this problem.

Answer:

{}, {s} {q} {l} {r} {s,q} {s, l} {s, r} {q, l} {q, r} {l, r} {s, q, l} {s,l,r} {s, q, r} {q, l, r} {s, q, l, r}