Evaluate by using the order of operations: 2+3(5-3)^2÷12

Hlp me pls identify inverse operations

Anika [276]

Inverse operation is an operation that undoes what was done by the previous operation. Take this simple addition problem 5+2=7. If we want to reverse the addition, we just subtract 7-2=5 and we are back where we started. You do the same for multiplication and division.

3 0

3 years ago

Quadrilateral ABCD is shown. If m angles?

artcher [175]

Answer: m∠BCE = 63°

m∠BAD = 76°

Step-by-step explanation:

Problem 1:

Given

m∠BEC = 90°

m∠EBC = 27°

Total = 117°

Solution:

Subtract 180° - 117° = m∠BCE = 63°

Problem 2:

Given

m∠ADE = 52°

m∠ABE = 52°

Total = 104°

Solution:

Subtract 180° - 104° = m∠BAD = 76°

6 0

3 years ago

What is the volume of the prism that can be constructed from this net? units3

kirza4 [7]

3×9=27
27×7=189 units³

I hope this helped!!!:)

4 0

3 years ago

Read 2 more answers

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$tan(\alpha)=cot(\beta)$

Find the value of the length side adjacent to the angle alpha

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

6 0

3 years ago

Express the product in simplest form.

Romashka [77]

$\frac{8}{2x+8} * \frac{ x^{2}-16 }{4}$

1. Cross cancel
$\frac{4}{2x+8}* \frac{ x^{2} -16}{1}$

2. Multiply numerators
4*x²-16 = 4x²-64

3. Multiply denominators
2x+8*1 = 2x+8

$\frac{4 x^{2} -64}{2x+8}$

4. Divide
$\frac{4 x^{2} -64}{2x+8} = 2x-8$

5. Simplify
2(x-4)

Answer is 2(x - 4)

4 0

3 years ago