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3 years ago

10

When comparing the graphs of y = x and y > x, what is a difference in the graphs?

1 answer:

mars1129 [50]3 years ago

8 0

I donnt think that there is a solution, but could be wrong

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A bag contains yellow marbles in red marbles 65 in total. The number of yellow marbles is five more than four times the number o

antoniya [11.8K]

Answer: there would be 52 yellow marbles

Step-by-step explanation:

Ok we know that there is 65 marbles in total? Right? So then we’d take 65 and think, what could we divide this by and start guessing and estimating numbers. (I guessed 5, one because it’s divisible by 65, two there is 5 times more yellow marbles than red marbles.) So 5 divided by 65 is 13. But, in order to find the amount of yellow marbles, we need to subtract the whole number by the amount of red marbles. (So 65-13 =?) if you subtract that you get 52 yellow marbles. :)

7 0

3 years ago

Given the piece wise function shown below, select all of the statements that are true

allsm [11]

C and d

when x = 2 , f(x) = x^2 = 4
and
when x = 1 f(x) = 5

8 0

3 years ago

Photo A, B, C are all square photos. The area of Photo C is the same as that of a rectangular photo whose length is the side len

motikmotik

Answer:

A

Step-by-step explanation:

because all off the porf on it was about a

5 0

3 years ago

Given points A (0,0), B (4,0), C (0,2), D (0,0), E (2,0), and F (0,1), which of the following proves that

damaskus [11]

Similar triangles may or may not have equal side lengths

Triangles ABC and DEF are similar by SSS theorem

<h3>How to determine the similarity statement</h3>

The coordinates of the two triangles are:

A (0,0), B (4,0), C (0,2), D (0,0), E (2,0), and F (0,1)

Calculate the lengths of both triangles using the following distance formula

$d = \sqrt{(x_2 -x_1)^2 + (y_2 -y_1)^2}$

For triangle ABC, we have:

$AB = \sqrt{(0 -4)^2 + (0 -0)^2} = 4$

$BC = \sqrt{(4 -0)^2 + (0 -2)^2} = 2\sqrt 5$

$CA = \sqrt{(0 -0)^2 + (0 -2)^2} = 2$

For triangle DEF, we have:

$DE = \sqrt{(0 -2)^2 + (0 -0)^2} = 2$

$EF = \sqrt{(2 -0)^2 + (0 -1)^2} = \sqrt 5$

$FD = \sqrt{(0 -0)^2 + (1 -0)^2} = 1$

Divide corresponding sides of the triangles to calculate the scale factor k

$k = \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD}$

This gives

$k = \frac{4}{2} = \frac{2\sqrt 5}{\sqrt 5} = \frac{2}{1}$

Evaluate the quotients

$k = 2 = 2 = 2$

Since the quotients are equal, then the triangles are similar

Hence, triangles ABC and DEF are similar by SSS theorem

Read more about similar triangles at:

brainly.com/question/14285697

3 0

2 years ago

Sixty percent of the students at a certain school wear neither a ring nor a necklace. Twenty percent wear a ring and 30 percent

Reptile [31]

Answer:

0.4

Step-by-step explanation:

Given

60 % wear neither ring nor a necklace

20 % wear a ring

30 % wear necklace

This question can be Solved by using Venn diagram

If one person is choosen randomly among the given student the probability that this student is wearing a ring or necklace is

$P\left ( wear \ ring\ or \ necklace )+P\left ( neither\ ring\ or\ necklace)=1$

$P\left ( wear \ ring\ or\ necklace )=1-0.6=0.4$

The sum of probabilty is equal to 1 because it completes the set

Therefore the required probabilty is 0.4

7 0

3 years ago