HELP PLS!!!!!!!!!!!!!!!!!!!!!

On your geometry test you have two triangles: AABC and AMNO. You are told that ZALM and that ZB

Dmitrij [34]

Answer:

b

Step-by-step explanation:

i took the test

4 0

3 years ago

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A family paid $30600 as a down payment for a home. If this represents %17 of the price of the home, find the price of the hom

snow_lady [41]

Answer:

180000

Step-by-step explanation:

30600/17=1800

1800•100=180000

7 0

3 years ago

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Leticia simplified an expression. Her work is shown below.

victus00 [196]

Leticia made a mistake in her step 2. She did subtraction before multiplication.

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<h3>How to simplify an expression?</h3>

The expression can be simplified as follows;

4.5 ÷ 0.25 + 2.5 - 0.75 × 8

Hence, using Pemdas rule,

4.5 ÷ 0.25 + 2.5 - 0.75 × 8

18 + 2.5 - 0.75 × 8

18 + 2.5 - 6 = 14.5

Therefore, she made mistake in her step 2 because she was suppose to solve the multiplication before subtraction.

learn more on expression here: brainly.com/question/13545210

#SPJ1

4 0

2 years ago

If G is the circumcenter of triangle ABC, find CG and FB. Round to the nearest tenth if necessary.

Stolb23 [73]

9514 1404 393

Answer:

CG = 19

FB = 14.7

Step-by-step explanation:

The circumcenter is equidistant from the triangle vertices, so GA = GB = GC.

The length of GA is given, so we have ...

GA = 19 = CG

__

Then the Pythagorean theorem can be used to find FB.

FB² +FG² = GB²

FB² = GB² -FG² = 19² -12² = 217

FB = √217 ≈ 14.7

3 0

3 years ago

In order to evaluate 7 sec(θ) dθ, multiply the integrand by sec(θ) + tan(θ) sec(θ) + tan(θ) . 7 sec(θ) dθ = 7 sec(θ) sec(θ) + ta

Maurinko [17]

Answer:

$\int {7 \sec(\theta) } \, d\theta = 7\ln(\sec(\theta) + \tan(\theta)) + c$

Step-by-step explanation:

The question is not properly formatted. However, the integral of $\int {7 \sec(\theta) } \, d\theta$ is as follows:

$\int {7 \sec(\theta) } \, d\theta$

Remove constant 7 out of the integrand

$\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) } \, d\theta$

Multiply by 1

$\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) * 1} \, d\theta$

Express 1 as: $\frac{\sec(\theta) + \tan(\theta) }{\sec(\theta) + \tan(\theta)}$

$\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) * \frac{\sec(\theta) + \tan(\theta) }{\sec(\theta) + \tan(\theta)}} \, d\theta$

Expand

$\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{\sec^2(\theta) + \sec(\theta)\tan(\theta) }{\sec(\theta) + \tan(\theta)}} \, d\theta$

Let

$u = \sec(\theta) + \tan(\theta)$

Differentiate

$\frac{du}{d\theta} = \sec(\theta)\tan(\theta) + sec^2(\theta)$

Make $d\theta$ the subject

$d\theta = \frac{du}{\sec(\theta)\tan(\theta) + sec^2(\theta)}$

So, we have:

$\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{\sec^2(\theta) + \sec(\theta)\tan(\theta) }{u}} \,* \frac{du}{\sec(\theta)\tan(\theta) + sec^2(\theta)}$

Cancel out $\sec(\theta)\tan(\theta) + sec^2(\theta)$

$\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{1}{u}} \,du}}$

Integrate

$\int {7 \sec(\theta) } \, d\theta = 7\ln(u) + c$

Recall that: $u = \sec(\theta) + \tan(\theta)$

$\int {7 \sec(\theta) } \, d\theta = 7\ln(\sec(\theta) + \tan(\theta)) + c$

8 0

3 years ago