A:1 hour 37 minutes B:2 hours 37 minutes C:2 hours 23 minutes D:1 hour 23 minutes

Janice is roller skating at a constant speed away from her house. The first half of the street is flat, but the second half has

Zolol [24]

Answer:

It's Z.

Step-by-step explanation:

Because the road is straight at first, then it gets steep.

which means that the road goes down,

kinda like a drop off.

8 0

3 years ago

Read 2 more answers

What is the distance, in units, between (3, -5) and (8, 7) in the coordinate plane?

Rufina [12.5K]

Yes, 13 is correct
If you draw the right triangle, the two legs will be 5 and 12.
5 and 12 are parts of a Pythagorean triple. 13 completes the triplet.

4 0

3 years ago

A system of equations is given below. For what value(s) of x is f(x)=g(x) ?

NARA [144]

$\large{f(x) = g(x)} \\ \large{ {x}^{2} + 3x = x + 3}$

Since we are solving the quadratic equation because the highest degree in the equation is second. We arrange in the form of ax²+bx+c = 0.

$\large{ {x}^{2} + 3x - x - 3 = 0}$

Combine like terms.

$\large{ {x}^{2} + 2x - 3 = 0}$

Solve the equation by factoring.

$\large{(x + 3)(x - 1) = 0} \\ \large{x = - 3,1}$

Hence the values of x that make f(x) = g(x) are -3 and 1.

Answer

x = -3,1

Let me know if you have any doubts!

3 0

3 years ago

Please I need 1,2, and 4

Virty [35]

Answer:what grade are you in?

Step-by-step explanation:

7 0

3 years ago

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$tan(\alpha)=cot(\beta)$

Find the value of the length side adjacent to the angle alpha

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

6 0

3 years ago