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3 years ago

Enter your answer and show all the steps that you use to solve this problem.

Mathematics

1 answer:

avanturin [10]3 years ago

3 0

9514 1404 393

Answer:

∠B = 41°

Step-by-step explanation:

Step 1: Recognize that ∠B and ∠A are "remote interior angles" with respect to the exterior angle at C. Hence their sum will be equal to the measure of the exterior angle.

Step 2. Compute angle B as the difference between the exterior angle at C and the interior angle at A:

113° - 72° = ∠B = 41°

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Evaluate the expression, if x = 9, y = 3, and z = 5. z2 • (x-y)

Bad White [126]

$\huge\text{Hey there!}$

$\mathsf{z^2 \times (x - y)}$

$\mathsf{= 5^2 \times (9 - 3)}$

$\mathsf{= 5 \times 5 \times (9 - 3)}$

$\mathsf{= \bold{25}\times(9 - 3)}$

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elena55 [62]

Answer:

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Step-by-step explanation:

Because it is a 90 degree angle, you subtract the 29 from 90 and you get 61. You can also put in in this equation and arrive at the same conclusion.

90=X+29

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One angle of an isosceles triangle measures 110°. Which other angles could be in that isosceles triangle? Choose all that apply.

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Step-by-step explanation:

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3 years ago

Write the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10).

Serjik [45]

Answer:

$\displaystyle f(x)=x^2+2x+2$

Step-by-step explanation:

System Of Linear Equations

In this problem, we'll need to solve a 3x3 system of linear equations because we have three unknowns and three conditions.

We are required to find the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10)

The general quadratic function can be written as

$\displaystyle f(x)=ax^2+bx+c$

We need to find the values of a,b, and c. Let's use the first condition, i.e. f(-1)=1

$\displaystyle f(-1)=a(-1)^2+b(-1)+c$

$\displaystyle f(-1)=a-b+c$

$\displaystyle a-b+c=1.....[eq\ 1]$

Now we use the second condition f(1)=5

$\displaystyle f(1)=a(1)^2+b(1)+c$

$\displaystyle f(1)=a+b+c$

$\displaystyle a+b+c=5.......[eq\ 2]$

Finally, we use the third condition f(2)=10

$\displaystyle f(2)=a(2)^2+b(2)+c$

$\displaystyle f(2)=4a+2b+c$

$\displaystyle 4a+2b+c=10....[eq\ 3]$

We put together eq 1, eq 2, and eq 3 to form the system

$\displaystyle \left\{\begin{matrix}a-b+c=1\\ a+b+c=5\\ 4a+2b+c=10\end{matrix}\right.$

Adding the first two equations we have

$\displaystyle 2a+2c=6$

$\displaystyle a+c=3$

And also

$\displaystyle b=2$

Using the above equation and the value of b in the third equation, we have

$\displaystyle \left\{\begin{matrix}a+c=3\\ 4a+c=6\end{matrix}\right.$

Subtracting the first equation from the second

$\displaystyle 3a=3$

$\displaystyle a=1$

And therefore

$\displaystyle c=2$

Now we have all the values, the quadratic function is

$\displaystyle \boxed{f(x)=x^2+2x+2}$

6 0

3 years ago