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3 years ago

In degrees, what is the measure of HMP? HELP WILL GIVE BRANLIEST+EXTRA POINTS

Mathematics

1 answer:

WITCHER [35]3 years ago

5 0

Answer:

The answer is 112 degrees.

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Negative number on a number line 2

Montano1993 [528]

I am pretty sure it is Point D

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2 years ago

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You are making a rectangular quilt with dimensions 48 inches by 60 inches. the fabric you are using costs $1.50 per square foot.

horsena [70]

We have to calculate the area of a rectangular quilt.
area of a rectangle=length x width
Data:
length=60 in=60 in(1 foot/12 in)=5 ft
width=40 in=40 in(1 foot/12 in)=3.33 ft

Area of the rectangula quilt=(5 in)(3.33 in)=16.67 ft²

Now, we have to calculate the cost of this quilt
$1.50-----------------1 ft²
x------------------------ 16.67 ft²

x=($1.5 * 16.67 ft²)/1 ft²=$25

Answer: the cost of the rectangular quilt was $25

5 0

3 years ago

A student stands on a bathroom scale in an elevator at rest on the 64 floor of a building. the scale reads 836n. as the elevator

strojnjashka [21]

At rest
mg=836 N
as g=9.8
m=836/9.8=85.30 kg
when going up,
mg+ma=935N
ma=935-836= 99N

a=99/85.30 = 1.16 m/s².
that is the acceleration of elevator
pls marks as brainliest answer

3 0

3 years ago

What is the antiderivative of sin^2(x)cos^2(x)?

marin [14]

Answer:

$\frac{x}{8}-\frac{\sin(4x)}{32}+C$

Step-by-step explanation:

[Most of the work here comes from manipulating the trig to make the term (integrand) integrable.]

Recall that we can express the squared trig functions in terms of cos(2x). That is,

$\cos(2x)=2\cos^2x-1 \\ \cos(2x)=1 - 2\sin^2x.$

And so inverting these,

$\cos^2x=\frac{1}{2} (1+\cos2x) \\ \sin^2x=\frac{1}{2} (1-\cos2x)$ .

Multiply them together to obtain an equivalent expression for sin^2(x)cos^2(x) in terms of cos(2x).

$\sin^2x \cdot \cos^2x =\frac{1}{2} (1-\cos2x) \cdot \frac{1}{2} (1+\cos2x) = \frac{1}{4}(1-\cos^2(2x)).$

Notice we have cos^2(2x) in the integrand now. We've made it worse! Let's try plugging back in to the first identity for cos^2(2x).

$\cos(2x)=2\cos^2x-1 \Rightarrow \cos(4x)=2\cos^2(2x)-1 \Rightarrow \cos^2(2x) = \frac{1}{2}(1+\cos(4x))$

So then,

$\sin^2x \cdot \cos^2x = \frac{1}{4}(1-\cos^2(2x)) = \frac{1}{4}(1-\frac{1}{2}(1+\cos(4x))) = \frac{1}{4}(1-\frac{1}{2}-\frac{1}{2}\cos(4x))=\frac{1}{8}(1-\cos(4x)).$

This is now integrable (phew),

$\int \sin^2x\cos^2x \ dx = \int \frac{1}{8}(1-\cos(4x)) \ dx = \frac{1}{8} \int (1-\cos(4x)) \ dx = \frac{1}{8}(x-\frac{1}{4}\sin(4x))+C.$

7 0

3 years ago

I don’t understand what this is asking

Kay [80]

Where’s the question

3 0

3 years ago