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Varvara68 [4.7K]

3 years ago

8

Someone pls help me I need help plsssss

1 answer:

guajiro [1.7K]3 years ago

4 0

Answer:

but I guess it can be all real numbers and shi te

Step-by-step explanation:

-3x-3 = -3(x+1)

-3x-3 = -3x-3 (which means -3(x+1) is basically putting x in evidence)

That way we have either:

-3x-3 = 0

-3x = 3

x = 3/-3

x = -1

or

-3(x+1) = 0

-3 cannot be 0 so:

x+1 = 0

x = -1

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The class sizes of the introductory psychology courses at a college are shown below 121,134,106,93,149,130,119,128 the college a

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Center : Mean Before the introduction of the new course, center = average(121,134,106,93,149,130,119,128) = 122.5 After the introduction of the new course, center = average(121,134,106,93,149,130,119,128,45) = 113.9 The center has moved to the left (if plotted in a graph) because of the low intake for the new course. Spread before introduction of the new course : Arrange the numbers in ascending order: (93, 106,119, 121), (128, 130,134, 149) Q1=median(93,106,119,121) = 112.5 Q3=median(128,130,134,149) = 132 Spread = Interquartile range = Q3-Q1 = 19.5 After addition of the new course,

(45,93, 106,119,) 121, (128, 130,134, 149)
Q1=median(45,93,106,119)=99.5
Q3=median (128, 130,134, 149)= 132
Spread = Interquartile range = 132-99.5 =32.5
We see that the spread has increased after the addition of the new course.

3 0

4 years ago

Read 2 more answers

What is the area of a regular hexagon with a distance from its center to a vertex of 1 cm? (Hint: A regular hexagon can be divid

devlian [24]

<h3>Answer:</h3><h3>Exact area = $\frac{3}{2}\sqrt{3}$ square cm</h3><h3>Approximate area = 2.598 square cm</h3>

=================================================

Work Shown:

s = side length of equilateral triangle = 1 cm

A = area of equilateral triangle with side length 's'

$A = \frac{\sqrt{3}}{4}*s^2$

$A = \frac{\sqrt{3}}{4}*1^2$

$A = \frac{\sqrt{3}}{4}$

This is just one of the 6 equilateral triangles (see diagram below)

Multiply by 6 to get the area of all 6 equilateral triangles, or the entire hexagonal area

$6*A = 6*\frac{\sqrt{3}}{4}$

$6A = \frac{3\sqrt{3}}{2}$

$6A \approx 2.598$

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3 years ago

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Answer: 170,000 kg m/s

Step-by-step explanation:

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3 years ago

Twice x, plus 8, is the sum as -10

Fofino [41]

Answer:

×=-9

Step-by-step explanation:

2x+8=-10

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3 0

4 years ago

Read 2 more answers

MNOP is a parallelogram. U is any point on side OP. Show that ar( MUN)= ar ( Δ △ PUM)+ar( UNO) △

bonufazy [111]

Answer:

The Proof is below.

Step-by-step explanation:

Given:

[]MNOP is a Parallelogram

U is any point on side OP

To Show:

ar(Δ MUN)= ar ( Δ PUM)+ar (Δ UNO)

Proof:

Theorem:

If a triangle and parallelogram are on the same base and have the same altitude, the area of the triangle will be half that of the parallelogram.

If they have same altitude, they will lie between the same parallels.

Hence the area of the triangle will be equal to half that of the parallelogram.

Area of Parallelogram will be twice of Area of Triangle

∴ $\textrm{Area of Parallelogram MNOP}=2\times (area\ of\ Triangle\ MUN)$ .............( 1 )

Also,

$\textrm{Area of Parallelogram MNOP}=Ar(MUN) + Ar(PUM)+ Ar(UNO)$

Substituting equation 1 we get

$2\times Ar(MUN)=Ar(MUN) + Ar(PUM)+ Ar(UNO)$

$2\times Ar(MUN)-Ar(MUN) = Ar(PUM)+ Ar(UNO)$

$Ar(MUN) = Ar(PUM)+ Ar(UNO)$ ...........Proved

8 0

4 years ago