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3 years ago

I need help ASAP !!!!!

Mathematics

1 answer:

mixas84 [53]3 years ago

6 0

Answer:

the y intercept is 3 , I'm positive it is 3

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Suppose you cut along a diagonal of a rhombus, a rectangle, or a square. Would you get two triangles that have the same shape an

stich3 [128]

I believe the answer to your question is

Yes, you would get two triangles that have the same shape and size

3 0

3 years ago

PLEASE HELP ME!!

Gnesinka [82]

First take the amount of drinks and popcorn needed then multiply the amount by the price then see if she has enough money when you finish the multiplying.

5 0

3 years ago

Use the integer tiles to evaluate the following expressions. 6 + 3 = 6 + (–4) = 6 + (–6) =

babunello [35]

Given:

The expressions are:

$6+3$

$6+(-4)$

$6+(-6)$

To find:

The value of given expression by using integer tiles.

Solution:

We have,

$6+3$

Here, both number are positive. When we add 6 and 3 positive integer tiles, we get 9 positive integer tiles as shown in the below figure. So,

$6+3=9$

Similarly,

$6+(-4)$

Here, 6 is positive and -4 is negative. It means we have 6 positive integer tiles and 4 negative integer tiles.

When we cancel the positive and negative integer tiles, we get 2 positive integer tiles as shown in the below figure. So,

$6+(-4)=2$

$6+(-6)$

Here, 6 is positive and -6 is negative. It means we have 6 positive integer tiles and 6 negative integer tiles.

When we cancel the positive and negative integer tiles, we get 0 integer tiles as shown in the below figure. So,

$6+(-6)=0$

Therefore, $6+3=9, 6+(-4)=2,6+(-6)=0$ .

8 0

2 years ago

Help pl0x, Algebra 1

djyliett [7]

$width=8+x+x=8+2x\\length=9+x+x=9+2x\\\\Area:A=width\cdot length\\\\A=(8+2x)(9+2x)=(8)(9)+(8)(2x)+(2x)(9)+(2x)(2x)\\\\=72+16x+18x+4x^2=4x^2+34x+72\\\\A=75ft^2\\therefore\\\\4x^2+34x+72=75\ \ \ \ |subtract\ 75\ from\ both\ sides\\\boxed{4x^2+34x-3=0}$

3 0

3 years ago

Weights of the vegetables in a field are normally distributed. From a sample Carl Cornfield determines the mean weight of a box

pantera1 [17]

To find the z-score for a weight of 196 oz., use

$z=\frac{x-\mu}{\sigma}=\frac{196-180}{8}=\frac{16}{8}=2$

A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.

1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.

7 0

3 years ago