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4 years ago

How can i work this problem out

Mathematics

1 answer:

ValentinkaMS [17]4 years ago

8 0

The form of the line is given by y=mx +b

Where b is the y intercept of the line

That is the point where the line cuts the y axis

So here b= 1

SoEquation is Y=1/3 x +(1)

Answer in place of question mark is 1

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Four minus the product of one and a number X

densk [106]

Answer:

4 - 1x

Step-by-step explanation:

This could also just be written as 4-x because one times anything is itself, so the one is kinda unnecessary but yup

6 0

3 years ago

Read 2 more answers

HELP PLEASE >>>>>>>>>>>>>>>>>>>>>>

Damm [24]

Answer:

A. $\frac{2}{8}$

B. $\frac{6}{6}$

C. $\frac{8}{4}$

Step-by-step explanation:

For A pick a fraction that is less than a whole number.

For B pick a fraction that is a whole number.

For C pick a fraction that is greater than a whole number.

3 0

3 years ago

How do i solve for y.<br> 2x^2 = 6 + 4x + y

rusak2 [61]

Answer:

you solve for x then find y

Step-by-step explanation:

4 0

4 years ago

Which polar equation represents an ellipse?

tatuchka [14]

Answer:

$r=\frac{1}{3+2cos\theta}$

Step-by-step explanation:

Let us write the equations in standard form:

$r=\frac{1}{3+2cos\theta} \implies r=\frac{\frac{1}{3} }{1+\frac{2}{3}\cos \theta }$

We have

$e=\frac{2}{3}\:$ and

$ep=\frac{1}{3}$

Since the eccentricity of this conic is less than 1, the conic represents an ellipse.

The second equation is $r=\frac{3}{2+3\sin \theta}$ .

This is a hyperbola, because eccentricity is more than 1.

The third equation is $r=\frac{5}{2+2\sin \theta}$ .

This is a parabola, because eccentricity is 1.

The fourth equation is $r=\frac{2}{2-3\sin \theta}$ .

This is also a hyperbola, because eccentricity is more than 1.

5 0

4 years ago

NEED HELP PLEASE!

USPshnik [31]

Answer:

$(x + (3+5i))^2 = x^2 + 6x + (10x+30)i - 16$

Step-by-step explanation:

Complex numbers identity:

The complex numbers identity is:

$i^2 = -1$

Square of the sum:

$(x + y)^2 = x^2 + 2xy + y^2$

In this question:

$(x + (3+5i))^2$

$x^2 + 2x(3+5i) + (3+5i)^2$

$x^2 + 6x + 10xi + 9 + 30i + 25i^2$

$x^2 + 6x + 9 + (10x+30)i - 25$

$x^2 + 6x + (10x+30)i - 16$

$(x + (3+5i))^2 = x^2 + 6x + (10x+30)i - 16$

5 0

3 years ago