The coordinates of a square are (4,4),(4,8),(8,8) and (?,?)

Mathematics

2 answers:

neonofarm [45]3 years ago

7 0

It would be (8,4) to complete the square. Try graphing it to prove it!

Anna11 [10]3 years ago

3 0

Answer revealed in the attachment below.

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Which expressions are equivalent to 4^-2 times 7^-2

Harlamova29_29 [7]

Step-by-step explanation:

$= {4}^{ - 2} \times {7}^{ - 2}$

$= {(4 \times 7)}^{ - 2}$

$= {28}^{ - 2}$

$= \frac{1}{ {28}^{2} }$

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3 years ago

2•5(16/2-2•3)-2<br> Simplified

Artyom0805 [142]

Answer: 18

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3 years ago

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tamaranim1 [39]

$3tan^{2} \theta +7sec\theta=3$
First I converted the equation terms into sine and cosine.
$tan^{2}\theta = \frac{sin^{2}\theta}{cos^{2}\theta}$ and $sec\theta= \frac{1}{cos\theta}$
Substitution:
$\frac{3sin^2\theta}{cos^2\theta} + \frac{7}{cos\theta} =3$
Common Denominator Created:
$\frac{3sin^2\theta}{cos^2\theta} + \frac{7cos\theta}{cos^2\theta} =3$
Multiply each term by the LCD:
$3sin^2\theta+7cos\theta=3cos^2\theta$
Substitution: Recall ⇒ $sin^2\theta =1-cos^2\theta$
$3(1-cos^2\theta)+7cos\theta=3cos^2\theta$
Distribute and collect all terms on one side:
$6cos^2\theta-7cos\theta-3=0$
Factor and set each factor equal to 0:
$(2cos\theta-3)(3cos\theta+1)=0$
$2cos\theta-3=0$ ⇒ $theta=cos^{-1} \frac{3}{2}$
$3cos\theta+1=0$ ⇒ $theta=cos^{-1} \frac{-1}{3}$
The 2nd factor provides only possible answer 109.5 degrees

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3 years ago

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forsale [732]

Answer:

$\frac{31x}{30}$