All categories

3 years ago

Help pls 40 points with review 3

Mathematics

1 answer:

Marianna [84]3 years ago

6 0

Answer:

Rectangle A is not a scale drawing.

Rectangle B is a scale drawing.

Step-by-step explanation:

A scale drawing involves ratios.

Rectangle A:

10cm is double the original 5cm giving us a 1 : 2 ratio.

However, 25cm is <em>not</em> double the original 20cm. This means that this rectangle is not to scale of the given rectangle.

Rectangle B:

10cm is double the original 5cm giving us a 1 : 2 ratio once again.

This time however, 40cm is double the original 20cm. This means that this rectangle is a scale drawing of the given rectangle.

~Hope this Helps!~

You might be interested in

Following the pattern in squaring a binomial, fill-in the missing term. I need fast, please

melisa1 [442]

Answer:

Step-by-step explanation:

im not completely sure

8 0

3 years ago

Ava and Maria went shopping together. Eva spent $20 more than Maria. Together they spent $209.84. Write and solve an equation to

ad-work [718]

Answer:209.84 - 20.00 = 189.84 / 2 = 94.92

94.92 + 20.00 = 114.92 Eva spent and Martha spent $94.92

Step-by-step explanation:

5 0

3 years ago

HELP!! What is the approximate measure of angle x in the triangle shown?

Alex Ar [27]

<h3>Answer: D) 130.5 degrees</h3>

=================================================

Work Shown:

$c^2 = a^2 + b^2 - 2*a*b*\cos(C)\\\\10^2 = 5^2 + 6^2 - 2*5*6*\cos(x)\\\\100 = 25 + 36 - 60*\cos(x)\\\\100 = 61 - 60*\cos(x)\\\\100 - 61 = - 60*\cos(x)\\\\39 = - 60*\cos(x)\\\\\cos(x) = \frac{39}{-60}\\\\\cos(x) = -0.65\\\\x = \arccos(-0.65)\\\\x \approx 130.5416\\\\x \approx 130.5\\\\$

Note: I used the law of cosines. Make sure your calculator is in degree mode.

6 0

3 years ago

What Val are restricted from the domain of

djverab [1.8K]

<u>Given</u>:

The given expression is $\frac{\left(4 x^{2}-4 x+1\right)}{(2 x-1)^{2}}$

We need to determine the values for which the domain is restricted.

<u>Restricted values:</u>

Let us determine the values restricted from the domain.

To determine the restricted values from the domain, let us set the denominator the function not equal to zero.

Thus, we have;

$(2x-1)^2\neq 0$

Taking square root on both sides, we get;

$2x-1\neq 0$

$2x\neq 1$

$x\neq \frac{1}{2}$

Thus, the restricted value from the domain is $x\neq \frac{1}{2}$

Hence, Option A is the correct answer.

7 0

3 years ago

Points M, N, and P are respectively the midpoints of sides AC , BC , and AB of △ABC. Prove that the area of △MNP is on fourth of

Hunter-Best [27]

Answer:

The area of △MNP is one fourth of the area of △ABC.

Step-by-step explanation:

It is given that the points M, N, and P are the midpoints of sides AC, BC and AB respectively. It means AC, BC and AB are median of the triangle ABC.

Median divides the area of a triangle in two equal parts.

Since the points M, N, and P are the midpoints of sides AC, BC and AB respectively, therefore MN, NP and MP are midsegments of the triangle.

Midsegments are the line segment which are connecting the midpoints of tro sides and parallel to third side. According to midpoint theorem the length of midsegment is half of length of third side.

Since MN, NP and MP are midsegments of the triangle, therefore the length of these sides are half of AB, AC and BC respectively. In triangle ABC and MNP corresponding side are proportional.

$\triangle ABC \sim \triangle NMP$