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3 years ago

7

I need help!! due in 8.25 at 11pm!! 15 points!!

2 answers:

Lyrx [107]3 years ago

6 0

Answer:

$\huge \boxed{y=-\frac{3}{4}x+6 }$

Step-by-step explanation:

Two points are (0, 6) and (8, 0)

Finding the slope.

m = (y2-y1)/(x2-x1)

m = (0-6)/(8-0)

m = -6/8

m = -3/4

The y-intercept is 6.

y = mx + b

m = slope

b = y-intercept

y = -3/4x + 6

11Alexandr11 [23.1K]3 years ago

3 0

Answer:

Hey there!

The equation is y=-3/4x+6. (Note the 3/4 is negative because the line has a negative slope.

Let me know if this helps :)

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F(x)=|3x+5|+6<br> g(x)=7<br> find (f+g) (x).

horrorfan [7]

Answer:

$(f+g) (x)=\mid 3x+5\mid+13$

Step-by-step explanation:

<u>Operation of Functions</u>

Given:

$f(x)=\mid 3x+5\mid+6$

$g(x)=7$

The sum of (f+g) (x) is:

$(f+g) (x)=\mid 3x+5\mid+6+7$

We cannot operate with the expression inside the absolute bars, thus:

$\boxed{(f+g) (x)=\mid 3x+5\mid+13}$

8 0

3 years ago

Other expressions for 19 times 25 + 19 times 75

Vladimir [108]

Answer:

1,900

Step-by-step explanation:

4 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

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

<em>Find the value of the length side adjacent to the angle alpha</em>

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

Original price 89.00 mark down 33%

denpristay [2]

You will mark down the product $29.37. So it would be 89.00-29.37=59.63

5 0

3 years ago

Katie has taken 5 math tests this year. On the first and second tests, she scored a 75. The third test she scored 10 points more

Zigmanuir [339]

75+75+85+95+90 = 420
420/5 = 84

6 0

3 years ago

Read 2 more answers