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3 years ago

15

A portrait without its frame has a height two times it’s width (w) The width of the frame is 3 inches on all sides. What is the

area of the portrait and the frame? simplify your answer

1 answer:

aliina [53]3 years ago

6 0

Answer:I'm guessing 9×6? 54?

Step-by-step explanation:

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Which shape , if rotated 90 degrees will coincide within itself

Likurg_2 [28]

THE ANSWER IS A
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7 0

3 years ago

What is -4 x 2 + 3x = -2

Brrunno [24]

Answer:

lets goooo

-4×2+3x=-2

you multiply -4 by 2 as shown in the question

so -4×2+3x=-2

-8+3x=-2

3x = -2+8

3x =6

you divide both sides by the x variable

3x =6

3x/3=6/3

x=3

5 0

3 years ago

Jamal, usually gets 5.28 hours of sleep each night. A teacher wants him to get more sleep. During an experiment, Jamal needs to

Nata [24]

Answer:

He will need to sleep 5.77 hours

Step-by-step explanation:

The first night, he slept 3 percent more than the previous night, meaning 5.28 + (5.28 x 3/100) = 5.28 + 0.1584 = 5.4384

The second night, he will need to sleep 3 percent more than the previous night, meaning 5.4384 + (5.4384 x 3/100) = 5.4384 + 0.163152 = 5.601552

The third night, he will need to sleep 3 percent more than the previous night, meaning 5.601552 + (5.601552 x 3/100) = 5.601552 + 0.16804656 = 5.76959856 = 5.77

7 0

3 years ago

Need help question #1. Show steps please

SSSSS [86.1K]

Answer:

C

Step-by-step explanation:

We want to integrate:

$\displaystyle \int\frac{4x^4+3}{4x^5+15x+2}\,dx$

Notice that the expression in the denominator is quite similar to the expression in the numerator. So, we can try performing u-substitution. Let u be the function in the denominator. So:

$u=4x^5+15x+2$

By differentiating both sides with respect to x:

$\displaystyle \frac{du}{dx}=20x^4+15$

We can "multiply" both sides by dx:

$du=20x^4+15\,dx$

And divide both sides by 5:

$\displaystyle \frac{1}{5}\, du=4x^4+3\,dx$

Rewriting our original integral yields:

$\displaystyle \int \frac{1}{4x^5+15x+2}(4x^4+3\, dx)$

Substitute:

$\displaystyle =\int \frac{1}{u}\Big(\frac{1}{5} \, du\Big)$

Simplify:

$\displaystyle =\frac{1}{5}\int \frac{1}{u}\, du$

This is a common integral:

$\displaystyle =\frac{1}{5}\ln|u|$

Back-substitute. Of course, we need the constant of integration:

$\displaystyle =\frac{1}{5}\ln|4x^5+15x+2|+C$

Our answer is C.

4 0

3 years ago

Listed below are the overhead widths (in cm) of seals measured from photographs and the weights (in kg) of the seals. Construc

gladu [14]

Answer:

1.) scatter plot is attached below.

There is no sufficient evidence to support the claim that there is a linear correlation between overhead widths of seals from photographs and the weights of the seals.

Step-by-step explanation:

Given the data :

Overhead width : 7.2 7.5 9.7 9.3 8.7 8.2

Weight : 119 156 243 200 199 188

The linear correlation Coefficient, R = 0.946

Using the Pvalue calculator for R score ;

Pvalue = 0.0149

Since, Pvalue > Α ; WE fail to reject the Null and conclude that there is no sufficient evidence to support the claim that there is a linear correlation between overhead widths of seals from photographs and the weights of the seals.

4 0

3 years ago