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3 years ago

Choose the equation below that represents the line passing through the point (−3, −1) with a slope of 4. (1 point)

Mathematics

2 answers:

tangare [24]3 years ago

6 0

It would be the first one

stellarik [79]3 years ago

4 0

The equation of a line is written using the form y - y1 = m(x-x1), where m is the slope and y1 and x1 are points on the line.

You are given the point (-3,-1) which means x1 = -3 and y1 = -1.

You are also told the slope is 4.

Replace those in the equation :

y - (-1) = 4(x-(-3)

simplify:

y+1 = 4(x+3)

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A parallelogram with a base of x units and a height of y units is changed

suter [353]

Answer:

868 (third option)

Step-by-step explanation:

we know that the area of original is $xy=31$ .

the new parallelogram will has area $4x\times 7y = 28xy=28\times 31=868$

6 0

3 years ago

Please help :c the first word is find

xeze [42]

Answer:

The values of $r_{2}$ and $\alpha_{2}$ are 2 and 150º.

Step-by-step explanation:

The complete statement is:

Find $\alpha_{2}$ and $r_{2}$ such that $\sin \theta - \sqrt{3}\cdot \cos \theta = r_{2}\cdot \cos (\theta - \alpha_{2})$ .

We proceed to use the following trigonometric identity:

$\cos (\theta - \alpha_{2}) = \cos \theta \cdot \cos \alpha_{2} +\sin \theta \cdot \sin \alpha_{2}$ (1)

$\sin \theta -\sqrt{3}\cdot \cos \theta = r_{2}\cdot \cos \theta \cdot \cos \alpha_{2}+r_{2}\cdot \sin \theta \cdot \sin \alpha_{2}$

By direct comparison we derive these expressions:

$r_{2}\cdot \sin \alpha_{2} = 1$ (2)

$r_{2}\cdot \cos \alpha_{2} = -\sqrt{3}$ (3)

By dividing (2) by (3), we have the following formula:

$\tan \alpha_{2} = -\frac{1}{\sqrt{3}}$

$\tan \alpha_{2} = -\frac{\sqrt{3}}{3}$

The tangent function is negative at second and fourth quadrants. That is:

$\alpha_{2} = \tan^{-1} \left(-\frac{\sqrt{3}}{3} \right)$

There are at least two solutions:

$\alpha_{2,1} = 150^{\circ}$ , $\alpha_{2,2} = 330^{\circ}$

And the value of $r_{2}$ :

$r_{2}^{2}\cdot \sin^{2}\alpha_{2} + r_{2}^{2}\cdot \cos^{2}\alpha_{2} = 4$

$r_{2}^{2} = 4$

$r_{2} = 2$

The values of $r_{2}$ and $\alpha_{2}$ are 2 and 150º.

5 0

3 years ago

A quadratic function, f(x) = x2 + bx + 9, is such that there is only one real root. Which of the following are possible values o

natita [175]

Answer:

Step-by-step explanation:

Remember that we can use the discriminant to determine the amount of roots that a quadratic function has.

If the determinant equals 0, then we only have one real root.

Our function is given by:

$f(x)=x^2+bx+9$

Then the discriminant will be:

$\Delta = b^2-4(1)(9)=b^2-36$

We only have one real root, thus our discriminant must be 0:

$0=b^2-36$

Solve for b:

$b^2=36$

Thus:

$b=\pm 6$

The answer is both II and III.

The final answer, then, is C.

7 0

3 years ago

What is the fraction for 105/100

Maksim231197 [3]

Its actually a mixed number that is 1 1/20 since you can reduce it from 1 5/100

4 0

3 years ago

Read 2 more answers

What is the ratio 5 to 105 expressed as a fraction in simplest form?

EastWind [94]

You start 5/105
Divide the numerator and denominator both by five to get 1/21

3 0

3 years ago