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3 years ago

If y=-4/5x-2, when is the value of x when y=-9

Mathematics

1 answer:

eimsori [14]3 years ago

6 0

Answer:

-55/4

Step-by-step explanation:

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The perimeter of a triangle is 30 units. If it is dilated with respect to origin by a scale factor of 0.5, what is the perimeter

il63 [147K]

<span>Choices: 30 units ; 20 units ; 15 units ; 7.5 units

Given:
perimeter = 30 units
scale factor = 0.5

30 units * 0.50 = 15 units

The perimeter of the resulting triangle is 15 units.</span>

6 0

3 years ago

If 433 = 100% then what percent does 383 equal

Goryan [66]

Answer:

88.46%

Step-by-step explanation:

433 - 383 = 50

50 ÷ 433 = 0.1154

0.1154 x 100 = 11.54%

100% - 11.54% = 88.46%

<u>Check work:</u>

433 x 88.46% = 383

5 0

3 years ago

Which of the following is a solution of x2 + 4x + 25?

blondinia [14]

Solve for x:
x^2 + 4 x + 25 = 0 I ssume that's the notation.
Subtract 25 from both sides:
x^2 + 4 x = -25
Add 4 to both sides:
x^2 + 4 x + 4 = -21
Write the left hand side as a square:
(x + 2)^2 = -21
Take the square root of both sides:
x + 2 = i sqrt(21) or x + 2 = -i sqrt(21)
Subtract 2 from both sides:
x = i sqrt(21) - 2 or x + 2 = -i sqrt(21)
Subtract 2 from both sides:
Answer: x = i sqrt(21) - 2 or x = -i sqrt(21) - 2

5 0

3 years ago

Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3

omeli [17]

$\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}$

$=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1$

$=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R$

3 0

3 years ago

If f(x) = 10 + 50x, solve for x when f(x) = 310.

weqwewe [10]

x = 6

given f(x) = 310, we obtain the equation

10 + 50x = 310 ( subtract 10 from both sides )

50x = 300 ( divide both sides by 50 )

x = $\frac{300}{50}$ = 6

5 0

3 years ago