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2 years ago

-3(k-8)-(k+5)=23 please explain how to do this

Mathematics

2 answers:

VMariaS [17]2 years ago

6 0

Answer:

k = -1

Step-by-step explanation:

Use the distributive property to eliminate parentheses. That property tells you that the outside factor multiplies each term inside parentheses:

-3(k -8) -(k +5) = 23

-3k -3(-8) -k -(+5) = 23

-3k +24 -k -5 = 23 . . . . simplify

-4k +19 = 23 . . . . . . . . . collect terms

-4k = 4 . . . . . . . . . . . . . subtract 19 from both sides

k = -1 . . . . . . . . . . . . . . . divide both sides by -4

_____

The rule for manipulating equations is, "any operation performed on one side of the equation must also be performed on the other side." Operations include addition, subtraction, multiplication, division, application of a function. Using the distributive property and collecting terms are ways to simplify equations that don't affect the truth of the equal sign (if those are done correctly).

In general, you make use of inverse operations to undo the operations you don't want. The goal is to get the variable by itself on one side of the equal sign.

lora16 [44]2 years ago

3 0

Answer:

k = -1

Step-by-step explanation:

-3(k-8)-(k+5)=23

Distribute

-3k +24 -k-5 = 23

Combine like terms

-4k +19 = 23

Subtract 19 from each side

-4k +19-19 = 23-19

-4k = 4

Divide by -4

-4k/-4 = 4/-4

k = -1

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Answer:

1. (7,-9)

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Step-by-step explanation:

To find the vertex use the formula

$\frac{-b}{2a}$

First make the equation into a quadratic equation.

$(x-4)(x-10)\\x^{2} -10x-4x+40\\x^{2} -14x+40$

Now it is in the form

$ax^{2} +bx+c=0$

Now we can substitute the values

$\frac{-(-14)}{2(1)} \\\frac{14}{2} \\7$

Now we have 7 as our x value for the vertex we can substitute for y,

$x^{2} -14x+40=y\\(7)^{2} -14(7)+40=y\\49-98+4=y\\y=-9$

So the vertex for 1 is (7,-9)

To find the intercepts for 2 & 3, they basically already give it to you. You just need to find a value for x for x-4 that will equal it to 0 and another for x for x-10 that will equal it to 0.

$x-4=0\\x=4\\\\x-10=0\\x=10$

4 is slightly different but start the same,

$(x-3)(x+3)\\x^{2}+3x-3x-9\\x^{2}-9$

Here, c (9) just shows the graph move down 9, so the y intercept = 9.

5 is the same process as 1,

$2(x-3)(x+5)\\2(x^{2}+5x-3x-15)\\2(x^{2}+2x-15)\\2x^{2}+4x-30$

Vertex,

$\frac{-(4)}{2(2)} \\\frac{-4}{4} \\-1$

Substitute,

$2x^{2}+4x-30\\2(-1)^{2}+4(-1)-30\\2-4-30\\-32$

So the vertex is (-1,-32)

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Step-by-step explanation:

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