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Fiesta28 [93]
3 years ago
6

563-273 do you have to regroup

Mathematics
1 answer:
Marizza181 [45]3 years ago
8 0
Yes you do because you cannot subtract 7 from 6.
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The perimeter of a rectangle is 10x^2+4x+4 the width is 2x-1 what is the length
fiasKO [112]

Answer:

10/2x^2+2x+2=2x-1+y

y=5x^2+1

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
A pound of apples costs $1.98 how much will 0.75 pounds cost ?
PolarNik [594]

Given:

Cost of one pound of apples = $1.98

To find:

The cost of 0.75 pound of apples.

Solution:

We have,

Cost of one pound = $1.98

So, cost of 0.75 pound of apples = 0.75 × cost of one pound of apples

\text{Cost of 0.75 pound of apples}=0.75\times 1.98

\text{Cost of 0.75 pound of apples}=1.485

Therefore, the cost of 0.75 pound of apples is $1.485.

8 0
4 years ago
Factor the following expression. 3 2 + 17 − 28<br><br> Explain your steps.
4vir4ik [10]
3(2)+17-28
multiply 3(2)=6
6+17-28
work left to right
6+17=23
23-28=-5
8 0
3 years ago
Read 2 more answers
Please can someone help I don’t understand the question if it’s right I’ll give more plsssssesssssssssssddsd
Anuta_ua [19.1K]

Answer:

72 i think if correct pls mark brainliest

Step-by-step explanation:

6 0
3 years ago
Write the trigonometric expression in terms of sine and cosine, and then simplify. cot()/sin()-csc()
OLEGan [10]

Answer:

First, we know that:

cot(x) = cos(x)/sin(x)

csc(x) = 1/sin(x)

I can't know for sure what is the exact equation, so I will assume two cases.

The first case is if the equation is:

\frac{cot(x)}{sin(x)} - csc(x)

if we replace cot(x) and csc(x) we get:

\frac{cot(x)}{sin(x)} - csc(x) = \frac{cos(x)}{sin(x)} \frac{1}{sin(x)}  - \frac{1}{sin(x)}

Now let's we can rewrite this as:

\frac{cos(x)}{sin(x)} \frac{1}{sin(x)}  - \frac{1}{sin(x)} =\frac{cos(x)}{sin^2(x)} - \frac{1}{sin(x)}

\frac{cos(x)}{sin^2(x)}  - \frac{sin(x)}{sin^2(x)} = \frac{cos(x) - sin(x)}{sin^2(x)}

We can't simplify it more.

Second case:

If the initial equation was

\frac{cot(x)}{sin(x) - csc(x)}

Then if we replace cot(x) and csc(x)

\frac{cos(x)}{sin(x)}*\frac{1}{sin(x) - 1/sin(x)} = \frac{cos(x)}{sin(x)}*\frac{1}{sin^2(x)/sin(x) - 1/sin(x)}

This is equal to:

\frac{cos(x)}{sin(x)}*\frac{sin(x)}{sin^2(x) - 1}

And we know that:

sin^2(x) + cos^2(x) = 1

Then:

sin^2(x) - 1 = -cos^2(x)

So we can replace that in our equation:

\frac{cos(x)}{sin(x)}*\frac{sin(x)}{sin^2(x) - 1} = \frac{cos(x)}{sin(x)}*\frac{sin(x)}{-cos^2(x)} = -\frac{cos(x)}{cos^2(x)}*\frac{sin(x)}{sin(x)}  = - \frac{1}{cos(x)}

5 0
3 years ago
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