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3 years ago

Help pls urgent :) plssss

Mathematics

1 answer:

Ket [755]3 years ago

7 0

Answer:

Fraction form: 1/8

Decimal form: 0.125

Step-by-step explanation:

Slope of line = [-6 - (-8)]/[8 - ( -8)]

= [-6 + 8]/[8 + 8]

= 2/16

= 1/8

= 0.125

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Sara is completing the square to find the maximum or minimum value of the function f(x) = (2 - x) (5 + x). What is the first ste

DanielleElmas [232]

Definitely multiply out the given factors:

f(x) = 10 + 2x - 5x - x^2, or f(x) = -x^2 - 3x + 10
Find the derivative: f '(x) = -2x - 3
Set the deriv. = to 0 and solve for x: -2x = 3, and x = -3/2
This x = -3/2 is the x-coordinate of the max value. The y-coord. is

f(-3/2) = (-3/2)^2 - 3(-3/2) + 10 = 21.25

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3 years ago

Is (2, 4) a solution of 2x – 7y < -10? Choose 1 answer: A Yes B No

iren2701 [21]

Answer: NO DEFINITLY NOT

Step-by-step explanation:

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3 years ago

Read 2 more answers

Cost A=0.6489 take the inverse cosine of both sides

Stolb23 [73]

I am not quite sure what the question asks for,
But this is what i assume it wants:

Cos A= 0.6489
In this given one, we basically find the size of the angle A
we do cosine inverse on both sides to get the size of the angle A
$cos^{-1}$ : It looks like this in the calculator
$cos^{-1}$ × cos A= $cos^{-1}$ (0.6489)
( $cos^{-1}$ and cos cancels out)
A= $cos^{-1}$ (0.6489)
A=49.54°
check:
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3 years ago

) f) 1 + cot²a = cosec²a

notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

$\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}$

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

$\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}$

Another identity is:

$\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}$

Therefore:

$\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}$

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

$\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}$