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mariarad [96]
4 years ago
7

Which ratios are in proportion to 2/5? Select all that apply.

Mathematics
2 answers:
kumpel [21]4 years ago
4 0
The answers are
b.
d
e.
musickatia [10]4 years ago
4 0
Answers are b (multiply by 3), d (multiply by 5), and e (multiply by 8).
You might be interested in
Two numbers total 65 and have a difference of 33. What are the two numbers?
DerKrebs [107]

Answer:

32 and 33

Step-by-step explanation:

so , 65-33= 32

32+33 is 65.

6 0
4 years ago
a girl painted a rectangular-shaped portrait which is 10 inches long and 8 inches wide. if she trimmed 2/1/2 inches on both side
frutty [35]

Answer:

32 in^2

Step-by-step explanation:

8-2=6, 6-2=4. 4 inches wide

10-2=8. 8 Inches tall.

4*8=32

7 0
3 years ago
An FBI agent orders a block of ballistics gel. The gel weighs 54 pounds per cubic foot. What is the weight of the block of gel?
Pavlova-9 [17]

Answer:

It Should Be 32

Step-by-step explanation:

Because if you add the 20 6 and 6 it gives you 32 hope this helps

6 0
3 years ago
Read 2 more answers
Which of the following functions would best model the situation above?
amid [387]
Usually bacteria expansion is modeled with exponential function.
Absolute value is not the correct answer since it models linear growth. 
Logarithmic is not the correct answer since it has an asymptote in the direction of the x axis. 
I will answer to this question "Polynomial" since it is likely to represent bacteria growth. 

3 0
4 years ago
Read 2 more answers
Be sure to answer all parts. List the evaluation points corresponding to the midpoint of each subinterval to three decimal place
gayaneshka [121]

Answer:

The Riemann Sum for \int\limits^5_4 {x^2+4} \, dx with n = 4 using midpoints is about 24.328125.

Step-by-step explanation:

We want to find the Riemann Sum for \int\limits^5_4 {x^2+4} \, dx with n = 4 using midpoints.

The Midpoint Sum uses the midpoints of a sub-interval:

\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+...+f\left(\frac{x_{n-2}+x_{n-1}}{2}\right)+f\left(\frac{x_{n-1}+x_{n}}{2}\right)\right)

where \Delta{x}=\frac{b-a}{n}

We know that a = 4, b = 5, n = 4.

Therefore, \Delta{x}=\frac{5-4}{4}=\frac{1}{4}

Divide the interval [4, 5] into n = 4 sub-intervals of length \Delta{x}=\frac{1}{4}

\left[4, \frac{17}{4}\right], \left[\frac{17}{4}, \frac{9}{2}\right], \left[\frac{9}{2}, \frac{19}{4}\right], \left[\frac{19}{4}, 5\right]

Now, we just evaluate the function at the midpoints:

f\left(\frac{x_{0}+x_{1}}{2}\right)=f\left(\frac{\left(4\right)+\left(\frac{17}{4}\right)}{2}\right)=f\left(\frac{33}{8}\right)=\frac{1345}{64}=21.015625

f\left(\frac{x_{1}+x_{2}}{2}\right)=f\left(\frac{\left(\frac{17}{4}\right)+\left(\frac{9}{2}\right)}{2}\right)=f\left(\frac{35}{8}\right)=\frac{1481}{64}=23.140625

f\left(\frac{x_{2}+x_{3}}{2}\right)=f\left(\frac{\left(\frac{9}{2}\right)+\left(\frac{19}{4}\right)}{2}\right)=f\left(\frac{37}{8}\right)=\frac{1625}{64}=25.390625

f\left(\frac{x_{3}+x_{4}}{2}\right)=f\left(\frac{\left(\frac{19}{4}\right)+\left(5\right)}{2}\right)=f\left(\frac{39}{8}\right)=\frac{1777}{64}=27.765625

Finally, use the Midpoint Sum formula

\frac{1}{4}(21.015625+23.140625+25.390625+27.765625)=24.328125

This is the sketch of the function and the approximating rectangles.

5 0
4 years ago
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