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2 years ago

Please answer this and thank you If DF=78, DE=5x-9, and EF=2x+10, find EF.

Mathematics

2 answers:

gregori [183]2 years ago

6 0

DE + EF = DF
5x - 9 + 2x + 10 = 78
7x+ 1 = 78
7x = 77
x = 77/7 = 11

EF = 2x + 10
EF = 2*11 + 10
EF = 22 + 10
EF = 32

AleksAgata [21]2 years ago

6 0

Question

If DF = 78, DE = 5x-9, and EF = 2x+10, find EF.

Answer

EF = 32

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The triangular prism shown has triangular _____face(s) and lateral face(s).

dmitriy555 [2]

Answer:

The triangular prism shown has 2 triangular face(s) and 3 lateral face(s).

The area of one triangular face is 7.5 mm²

The surface area of the triangular prism is 135 mm²

Step-by-step explanation:

A triangular face of the prism is the face that has a three sided figure while the lateral faces are faces with four-sided figure. As seen from the triangular prism, it has 2 triangular faces since it is made up of 2 right angled triangles and has 3 lateral faces since it is made up of 3 rectangles which are 4 sided each.

Area of one triangular face = $\frac{1}{2} * base * height$

Given base = 2.5 mm and height = 6 mm

Area of one triangular face = $\frac{1}{2} *2.5 * 6\\$

$= 3*2.5\\= 7.5mm^{2}$

Surface area of triangular prism = bh + pH

b= base of the triangle = 2.5 mm

h- height of the triangle = 6 mm

p= perimeter of the triangle = base + height + slant height = 2.5+6+6.5

p = 15 mm

H= height of the prism = 8mm

Surface area of triangular prism = 2.5(6)+15(8)

Surface area of triangular prism = 15 + 120

Surface area of triangular prism = 135mm²

5 0

2 years ago

Expand and simplify (k-3)^7

nikklg [1K]

There are a couple of ways you can solve this. You can write it out like
(K-3)(K-3)(K-3)(K-3)(K-3)(K-3)(K-3)
And pretty much use the foil method
Or you can use the binomial theorem (google it if you dont know)
The solution should be:
(K^7) - 21(k^6) + 189(k^5) - 945(k^4) + 2835(k^3) - 5103(k^2) + 5103k - 2187

3 0

2 years ago

Read 2 more answers

Consider the following formula used to estimate height, h, of a seedling after w weeks since planting:

svp [43]

Answer:

Option B. $w=5h-8$

Step-by-step explanation:

we have

$h=(\frac{1}{5})w+(\frac{8}{5})$

Solve for w

That means ----> isolate the variable w

Multiply by 5 both sides to remove the fraction

$5h=w+8$

subtract 8 both sides

$5h-8=w+8-8$

$5h-8=w$

Rewrite

$w=5h-8$

4 0

2 years ago

The combined average weight of an okapi and a llama is 450450450 kilograms. The average weight of 333 llamas is 190190190 kilogr

Kisachek [45]

I'm going to assume that you meant 450kg for the combined weight, 190kg more and 3 Llamas. I'm pretty sure Llamas and Okapis don't weigh 450450450kg (that's 993,073,252 pounds). :)

x= Okapi weight
y= Llama weight

EQUATIONS:
There are 2 equations to be written:

1) 450kg is equal to the weight of one Okapi and one Llama

450kg= x + y

2) The weight of 3 llamas is equal to the weight of one Okapi plus 190kg.

3y=190kg + x

STEP 1:
Solve for one variable in one equation and substitute the answer in the other equation.

450kg= x + y
Subtract y from both sides
450-y =x

STEP 2:
Substitute (450-y) in second equation in place of x to solve for y.

3y=190kg + x
3y=190 + (450-y)
3y=640 -y
add y to both sides

4y=640
divide both sides by 4

y=160kg Llama weight

STEP 3:
Substitute 160kg in either equation to solve for x.

3y=190kg + x
3(160)=190 + x
480=190 + x
Subtract both sides by 190

290= x
x= 290kg Okapi weight

CHECK:
3y=190kg + x
3(160)=190 + 290
480=480

Hope this helps! :)

8 0

2 years ago

The cost of 5 gallons of ice cream has a standard deviation of 8 dollars with a mean of 29 dollars during the summer. What is th

Feliz [49]

Answer:

97.74% probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 29, \sigma = 8, n = 92, s = \frac{8}{\sqrt{92}} = 0.8341$

What is the probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected?

This is the pvalue of Z when X = 29 + 1.9 = 30.9 subtracted by the pvalue of Z when X = 29 - 1.9 = 27.1. So

X = 30.9

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{30.9 - 29}{0.8341}$

$Z = 2.28$

$Z = 2.28$ has a pvalue of 0.9887

X = 27.1

$Z = \frac{X - \mu}{s}$

$Z = \frac{27.1 - 29}{0.8341}$

$Z = -2.28$

$Z = -2.28$ has a pvalue of 0.0113

0.9887 - 0.0113 = 0.9774

97.74% probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected

7 0

2 years ago