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3 years ago

On a number line, what is the unit distance between –11 and 3?

Mathematics

2 answers:

Gala2k [10]3 years ago

8 0

Answer:

Step-by-step explanation:

if you take 14 - 11 it equals 3 so that is a for sure hard 14

kumpel [21]3 years ago

3 0

The answer is 14

explanation answer: 14

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A letter is selected from the words in accommodation list all the possible outcomes

Luda [366]

Since a letter is selected from a c c o m m o d a t i o n

Possible outcome:

It could be any of the thirteen letters. There are thirteen possibilities.

a c c o m m o d a t i o n

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2 years ago

The dimensions of a wooden bench shaped like a rectangular prism are 6' x 3.5' x 2.5'. What is the volume of the bench in cubic

Dafna11 [192]

Answer:

Step-by-step explanation:

Volume of a rectangular prism

V=l*w*h

V=6*3.5*2.5=52.50 ft^3

8 0

2 years ago

1. Find the measure of the given angle. 156° L N ? M 124

Dominik [7]

Answer:

40°

Step-by-step explanation:

360-156-124=80

80/2=40

8 0

2 years ago

Solve for x. Your answer must be simplified. -3 < x/-4

Harman [31]

Answer:

x<12

Step-by-step explanation:

Let's solve your inequality step-by-step.

−3< x/ −4

Step 1: Simplify both sides of the inequality.

−3< −1/ 4 x

Step 2: Flip the equation.

−1/ 4 x>−3

Step 3: Multiply both sides by 4/(-1).

( 4/ −1 )*( −1/ 4 x)>( 4 /−1 )*(−3)

x<12

Answer:

x<12

4 0

2 years ago

Write the equation of a polynomial of degree 3, with zeros 1, 2 and -1 where f(0)=2

drek231 [11]

Answer:

The equation of a polynomial of degree 3, with zeros 1, 2 and -1 is $x^{3}-2 x^{2}-x+2=0$

Solution:

Given, the polynomial has degree 3 and roots as 1, 2, and -1. And f(0) = 2.

We have to find the equation of the above polynomial.

We know that, general equation of 3rd degree polynomial is

$F(x)=x^{3}-(a+b+c) x^{2}+(a b+b c+a c) x-a b c=0$

where a, b, c are roots of the polynomial.

Here in our problem, a = 1, b = 2, c = -1.

Substitute the above values in f(x)

$F(x)=x^{3}-(1+2+(-1)) x^{2}+(1 \times 2+2(-1)+1(-1)) x-1 \times 2 \times(-1)=0$

$\begin{array}{l}{\rightarrow x^{3}-(3-1) x^{2}+(2-2-1) x-(-2)=0} \\ {\rightarrow x^{3}-(2) x^{2}+(-1) x-(-2)=0} \\ {\rightarrow x^{3}-2 x^{2}-x+2=0}\end{array}$

So, the equation is $x^{3}-2 x^{2}-x+2=0$

Let us put x = 0 in f(x) to check whether our answer is correct or not.

$\mathrm{F}(0) \rightarrow 0^{3}-2(0)^{2}-0+2=2$

Hence, the equation of the polynomial is $x^{3}-2 x^{2}-x+2=0$

3 0

2 years ago