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3 years ago

How do you solve this equation

Mathematics

2 answers:

dsp733 years ago

5 0

First, you must get rid of the fractions. You can do that by multiplying that 15. Then, you get:
45x-12=-5x
Then combine the like terms

50x=12

x=12/50=6/25

uysha [10]3 years ago

5 0

Go to this website called cymath that then enter your equation or equations and it will help u GO NOWW!!!! :) :) :) ;) ;) =-)

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Please help this is important

Galina-37 [17]

Answer:

First blank - SSS congruent condition

Second blank - Congruent 

5 0

2 years ago

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36=2+(-12.4) please helpssss

Nikolay [14]

I believe the answer is 36= -10.4!
Sorry if I'm wrong though.

8 0

3 years ago

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Anna is draining water out of a 50 gallon rain barrel. The graph shows the relationship between time in minutes and the gallons

tatiyna

Answer:

6 < x ≤ 20

Step-by-step explanation:

First, let's use the points (0, 50) and (20, 0) to find the equation of the line.

m = Δy / Δx

m = (0 − 50) / (20 − 0)

m = -2.5

y = -2.5x + 50

Now find the time when y = 35.

35 = -2.5x + 50

-15 = -2.5x

x = 6

So there is less than 35 gallons in the barrel when 6 < x ≤ 20.

7 0

3 years ago

A variety of two types of snack packs are delivered to a store. The box plots compare the number of calories in each snack pack

adelina 88 [10]

OPTIONS:

A. The interquartile range of the trail mix data is greater than the range of the cracker data.

B. The value 70 is an outlier in the trail mix data.

C. The upper quartile of the trail mix data is equal to the maximum value of the cracker data.

D. The number of calories in the packs of trail mix have a greater variation than the number of calories in the packs of crackers.

Answer:

Step-by-step explanation:

With the given information about how the box plot looks like, let's examine each option to see if they are true or not.

Option A: "The interquartile range of the trail mix data is greater than the range of the cracker data."

The interquartile range of trail mix data = 105 - 90 = 15

Range of cracker data = 100 - 70 = 30

Option A is NOT TRUE.

Option B: "The value 70 is an outlier in the trail mix data."

This is NOT TRUE. There are not outliers as 70 is the minimum value if the ranges of the data set for the trail mix.

Option C: "The upper quartile of the trail mix data is equal to the maximum value of the cracker data."

Upper quartile of the trail mix data = 105

Max value of cracker data = 100

This statement is NOT TRUE.

Option D: "The number of calories in the packs of trail mix have a greater variation than the number of calories in the packs of crackers."

The greater the range value, the greater the variation. Thus,

Range value of the trail mix data = 115 - 70 = 45

Range value of the cracker data = 100 - 70 = 30

This is statement is correct because trail mix data have a greater range value, hence, it has a greater variation in the number of calories.

6 0

3 years ago

Read 2 more answers

A triangle has vertices at (-2,-3),(4, -3), and (3,5). What is the area of the triangle?

Bezzdna [24]

Answer:

this triangle is ABC with A(-2 ; -3), B(4; - 3) and C(3;5)

=> $AB=\sqrt{(4-(-2))^{2}+\sqrt{(-3 -(-3))^{2} } }=\sqrt{6^{2} }=6\\\\AC=\sqrt{(3-(-2))^{2}+(5-(-3))^{2} }=\sqrt{5^{2}+8^{2} }=\sqrt{89}\\\\BC =\sqrt{1^{2}+8^{2} }=\sqrt{65}$

using Heron theorem, we have:

$S=\sqrt{p(p-AB)(p-AC)(p-BC)}\\\\S=\sqrt{(\frac{6+\sqrt{89}+\sqrt{65} }{2})(\frac{\sqrt{89}+\sqrt{65}-6 }{2} )(\frac{6+\sqrt{65}-\sqrt{89} }{2})(\frac{6+\sqrt{89}-\sqrt{65} }{2}) }\\\\S=24 \\\\$

with S is the area of the triangle

$p=\frac{AB+AC+BC}{2}=\frac{6+\sqrt{89}+\sqrt{65} }{2}$

Step-by-step explanation:

5 0

3 years ago