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Zina [86]
3 years ago
9

Solve the inequality for x. –2x – 3 > 13 A) x < –5 B) x < –8 C) x > –5 D) x > –8

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
5 0

Answer:

B x<-8

Step-by-step explanation:

-2x -3> 13

-2x > 13+3

-2x >16

since you will divide both side by a negative number, the operator will change from > to <

-2x /-2 < 16/-2

x<-8

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Express 6 time the difference of 20 and 6 divide by 7 and simplify​
Stells [14]

Answer:

Here's how to reduce a fraction: Break down both the numerator (top number) and denominator (bottom number) into their prime factors. Cross out any common factors. Multiply the remaining numbers to get the reduced numerator and denominator.

6 0
2 years ago
Use the law of cosines to find the indicated missing side
max2010maxim [7]

Answer:

19.08

Step-by-step explanation:

c² = a² + b² - 2ab×cos(C)

a=10

b=22

C=60

cos(60) = 0.5

c² = 10² + 22² - 2×10×22×cos(60) =

= 100 + 484 - 20×22×0.5 = 584 - 20×11 = 584 - 220 =

= 364

c = sqrt(364) ≈ 19.08

7 0
3 years ago
a toy company spends $1,500 each day for factory expenses plus $8 per teddy bear which sells for $12 each. How many bears must t
Lunna [17]
1500÷8
1500÷12
subtract
6 0
3 years ago
Use Euler's Formula to find the missing number.<br>Faces: 20<br>Edges: 30<br>Vertices:​
adoni [48]

Answer:

V=12

Step-by-step explanation:

Faces: 20 Edges: 30

Write the Euler's formula for 3-dimensional figures

F+V=E+2

Substitute some variables for their known values

20+V=30+2

Add the numbers on the right side of the equation

20+v=32

Subtract 20 on both sides

20-20+V=32-20

Subtract

V=32-20

Number of Vertices

V=12

7 0
3 years ago
Consider the function below. f(x)= x^3 + 2x^2 - x - 2 plot the x and y intercepts of the function
dimulka [17.4K]

In the Figure below is shown the graph of this function. We have the following function:

f(x)=x^3+2x^2-x-2

The y-intercept occurs when x=0, so:

f(0)=(0)^3+2(0)^2-(0)-2=-2

Therefore, the y-intercept is the given by the point:

\boxed{(0,-2)}

From the figure we have three x-intercepts:

\boxed{P_{1}(-2,0)} \\ \boxed{P_{2}(-1,0)} \\ \boxed{P_{3}(1,0)}

So, the x-intercepts occur when y=0. Thus, proving this:

f(x)=x^3+2x^2-x-2 \\ \\ For \ P_{1}:\\ If \ x=-2, \ y=(-2)^3+2(-2)^2-(-2)-2=0 \\ \\ For \ P_{2}:\\ If \ x=-1, \ y=(-1)^3+2(-1)^2-(-1)-2=0 \\ \\ For \ P_{3}:\\ If \ x=1, \ y=(1)^3+2(1)^2-(1)-2=0

7 0
3 years ago
Read 2 more answers
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