All categories

3 years ago

What is the solution for -10 < x - 9?

Mathematics

1 answer:

enyata [817]3 years ago

4 0

-10 < x - 9

-10+9 < x - 9+9

-1<x, or
x> - 1

You might be interested in

Solve -52 = 4m for m

ZanzabumX [31]

Note the equal sign. What you do to one side, you do to the other.

Isolate the variable. Divide 4 from both sides

(-52)/4 = (4m)/4

m = -52/4

m = -13

-13 is your answer for m

hope this helps

8 0

3 years ago

Read 2 more answers

Give the equation of the line passing through the point (20,2)that is perpendicular to y=−4x.

Cerrena [4.2K]

Answer:

y-2 = 1/4(x-20) point slope form

y = 1/4x -3 slope intercept form

Step-by-step explanation:

y = -4x

The slope is -4

The slope that is perpendicular is the negative reciprocal

- (1/-4) = 1/4

The slope of our new line is 1/4

We have a point and the slope, so we can use point slope form

y-y1 = m(x-x1)

y-2 = 1/4(x-20)

If we want the line in slope intercept form, distribute

y-2 = 1/4x -5

Add 2 to each side

y-2+2 = 1/4x -5+2

y = 1/4x -3

5 0

3 years ago

The expression (secx + tanx)2 is the same as _____.

trapecia [35]

Answer:

The expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

Solution:

From question, given that $\bold{(\sec x+\tan x)^{2}}$

By using the trigonometric identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$ the above equation becomes,

$(\sec x+\tan x)^{2} = \sec ^{2} x+2 \sec x \tan x+\tan ^{2} x$

We know that $\sec x=\frac{1}{\cos x} ; \tan x=\frac{\sin x}{\cos x}$

$(\sec x+\tan x)^{2}=\frac{1}{\cos ^{2} x}+2 \frac{1}{\cos x} \frac{\sin x}{\cos x}+\frac{\sin ^{2} x}{\cos ^{2} x}$

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

On simplication we get

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

By using the trigonometric identity $\cos ^{2} x=1-\sin ^{2} x$ ,the above equation becomes

$=\frac{1+2 \sin x+\sin ^{2} x}{1-\sin ^{2} x}$

By using the trigonometric identity $(a+b)^{2}=a^{2}+2ab+b^{2}$

we get $1+2 \sin x+\sin ^{2} x=(1+\sin x)^{2}$

$=\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}$

$=\frac{(1+\sin x)(1+\sin x)}{1-\sin ^{2} x}$

By using the trigonometric identity $a^{2}-b^{2}=(a+b)(a-b)$ we get $1-\sin ^{2} x=(1+\sin x)(1-\sin x)$

$=\frac{(1+\sin x)(1+\sin x)}{(1+\sin x)(1-\sin x)}$

$= \frac{1+\sin x}{1-\sin x}$

Hence the expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

8 0

3 years ago

Solve the following equation algebraically 2x ^ 2 = 50

kipiarov [429]

Answer:

x=5

Step-by-step explanation:

$2 {x}^{2} = 50$

${x}^{2} = \frac{50}{2}$

${x}^{2} = 25$

$x = \sqrt{25}$

$x = 5$

7 0

3 years ago

Answers matching "The combined average weight of an okapi and a llama is 450 kilograms. The average weight of 3 llamas is 190 ki

kirill [66]

Let us say that:

x = weight of llamas

y = weight of okapi

From the problem, we can create the equations:

x + y = 450 --> 1

3 x = y + 190 --> 2

Rewriting equation 1:

x = 450 – y

From equation 2:

3 (450 – y) = y + 190

1350 – 3 y = y + 190

4 y = 1160

y = 290

From equation 1:

x = 450 – 290

x = 160

Therefore a llama weighs 160 kilograms while okapi weigh 290 kilograms on average.

5 0

4 years ago