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3 years ago

Last one guys I promise lol. btw helpp

Mathematics

1 answer:

Marianna [84]3 years ago

8 0

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Write the coordinates of the vertices after a reflection over the y-axis.

mart [117]

Answer:

A (10, -9)

B (0,-9)

C (0,-1)

D (10,-1)

Step-by-step explanation:

Negate the x-coordinates of each of the vertices

8 0

2 years ago

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Verify the trigonometric identity. convert the left side of the relationship to look like the right side.

Tomtit [17]

Answer:

Verified

Step-by-step explanation:

Use $\tan{x}=\frac{\sin{x}}{\cos{x}}$ , $\sin^2{x}+\cos^2{x}=1$ and $\sec{x}=\frac{1}{\cos{x}}$

LHS= $\tan^2{x}+1=\frac{\sin^2{x}}{\cos^2{x}} + 1=\frac{\sin^2{x}+\cos^2{x}}{\cos^2{x}}=\frac{1}{\cos^2{x}}=\sec^2{x}$ =RHS

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2 years ago

A stack of magazines is 4 2/5 inches high. The stack needs to fit into a box that is 2 1/8 inches high. How many inches too high

iren [92.7K]

Answer: Equation: 4 $\frac{2}{5}$ -2 $\frac{1}{8}$ ; estimate: a

little more than 2 inches; answer: 2 $\frac{11}{40}$

inches.

Step-by-step explanation:

6 0

3 years ago

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Gradient from 2 - 1/2y = 0

kirill115 [55]

Given :

2 - 1/2y = 0

To find :-

The gradient form .

Solution :-

The gradient form/ slope intercept form

y = mx + c ,

On converting ,

2 - 1/2y = 0
1/2y = 2
y = 2*2
y = 4
y = 0x + 4

4 0

2 years ago

Express p in terms of q if

Irina-Kira [14]

$p=x^2+\dfrac{1}{x^2}\\\\p=\dfrac{x^4}{x^2}+\dfrac{1}{x^2}\\\\p=\dfrac{x^4+1}{x^2}\qquad(*)$

$q=x+\dfrac{1}{x}\\\\q=\dfrac{x^2}{x}+\dfrac{1}{x}\\\\q=\dfrac{x^2+1}{x}\qquad\text{square both sides}\\\\q^2=\left(\dfrac{x^2+1}{x}\right)^2\\\\q^2=\dfrac{(x^2+1)^2}{x^2}\qquad\text{use}\ \ (a+b)^2=a^2+2ab+b^2\\\\q^2=\dfrac{(x^2)^2+2(x^2)(1)+1^2}{x^2}\\\\q^2=\dfrac{x^4+2x^2+1}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+\dfrac{2x^2}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+2\qquad\text{subtract 2 from both sides}\\\\q^2-2=\dfrac{x^4+1}{x^2}\\\\\text{From (*) we have}\\\\\boxed{p=q^2-2}$

8 0

3 years ago