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3 years ago

To find the y intercept of the graph of a linear equation we.

Mathematics

1 answer:

Lelechka [254]3 years ago

3 0

Answer:

Step-by-step explanation:

A y-intercept's x will always be zero, there is no movement along the horizontal axis. Since you are looking for a y-intercept, you will solve for y.

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HI! Help me please! HOMEWORKKKK UGHHH

avanturin [10]

Answer:

Step-by-step explanation:

if you multiply any number by a number lower than 1, then the value lowers. J is the only thing that has both numbers greater than one

6 0

3 years ago

Read 2 more answers

How did the tempicher change if at first it increased by 5% and then increased by 20 percent

tensa zangetsu [6.8K]

Answer:

Increasing a number by 5% and then by 20% is the same as increasing the original number by 26%.

Step-by-step explanation:

Take a number, x.

Now increase it by 5%.

1.05x

Now increase it by 20%.

1.2 * 1.05x = 1.26x

1.26x = 126% of x = 100% of x + 26% of x

100% of x is the same as x, so it is the same as the original x.

The increase is 26% of the original number.

Increasing a number by 5% and then by 20% is the same as increasing the original number by 26%.

3 0

3 years ago

Parallel lines always,sometimes,or never intersect?

STatiana [176]

They never intersect

8 0

3 years ago

Read 2 more answers

Thank you all for helping so far, mind if I ask again, haha

wariber [46]

Answer:

10 < 18 + 7

10 < 25

True yyes

cuz it is above 10

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

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

drek231 [11]

<u>Answer:</u>

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

<u>Solution:</u>

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

3 years ago