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lorasvet [3.4K]
3 years ago
12

5. – 3(1 + 6r) = 14 - 7​

Mathematics
1 answer:
MAVERICK [17]3 years ago
3 0

Answer:

r= -1.8

Step-by-step explanation:

Distribute

-3-18r=14-7

Combine like terms

-3-18r=7

add 3 to both sides

-18r=10

divide by -18

r=-1.8

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sandy is having a thanksgiving feast for her family of 24. she is deciding between these two options. Help please which one is b
valentina_108 [34]

Answer:

option b

Step-by-step explanation:

3 0
3 years ago
Use the quadratic formula to find both solutions to the quadratic equation given below x^2+6x=27
docker41 [41]

Answer:

x=3 or x=−9

Step-by-step explanation:

Step 1: Subtract 27 from both sides.

x2+6x−27=27−27

x2+6x−27=0

Step 2: Factor left side of equation.

(x−3)(x+9)=0

Step 3: Set factors equal to 0.

x−3=0 or x+9=0

x=3 or x=−9

Answer:

x=3 or x=−9

7 0
3 years ago
Read 2 more answers
Write an expression that is equivalent to -2/5(15 - 20d +50)<br> Explain please!:)
valina [46]

Answer:

-26 + 4/5d

Step-by-step explanation:

-2/5 ( 15 - 2d + 50)

= -2/5 ( 65 - 2d)

= -2/5 * 65 - (-2/5) * 2d

= -26 - (-4/5d)

= -26 + 4/5d

3 0
3 years ago
How can you use the measures of center and the range to compare two populations?
DochEvi [55]

The measure of center and range to be used to compare the population is the median and the inter quartile range.

<u>Explanation:</u>

You utilize the median and the interquartile range (IQR) to portray skewed distributions of information. To look at two populaces, utilize the mean and the MAD when the two populations are symmetric. Utilize the middle and the IQR when possibly one or the two circulations are slanted.

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8 0
3 years ago
The points A BC :(2, 2,1), :(1,1,3), :(2,0,5) − are the vertices of a right triangle. The radius of the sphere with center at th
ElenaW [278]

Answer:

r=1

Step-by-step explanation:

First we need to know the length of each side of the triangle, so we use the formula of the vector modulus:

|AB|= \sqrt{(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}+(b_{3}-a_{3})^{2}

By doing so, we find:

|AB|=\sqrt {6}\\|BC|=\sqrt {6}\\|AC|=2\sqrt {5}

With this we know that the triangle is not right, but, we assume the longest side as the hypotenuse of the problem.

As we have two equal sides, we know that the line between point |AB| and the center of the hypotenuse is perpendicular, therefore, we can calculate it using Pythagoras theorem:

|BC|^{2}=r^{2}+(\frac{|AC|}{2})^{2}\\\\r^{2}=|BC|^{2}-(\frac{|AC|}{2})^{2}\\\\r^{2}=(\sqrt{6})^{2}- (\frac{2\sqrt{5}}{2})^{2}\\\\r^{2}=6-5=1\\r=1

4 0
3 years ago
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