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slega [8]
3 years ago
8

Is 53 prime or composite

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
8 0
Composite number........
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The area of a rectangle is represented by x^2- 7x + 10. What are the missing side lengths of the rectangle?
ladessa [460]
Length = x - 2
Width = x - 5

Explanation:

Area = x^2 - 7x + 10
Area = (x - 2)(x - 5)
Area = Length x Width

Length = x - 2
Width = x - 5
4 0
2 years ago
-91=13n How would you work out this math problem
Igoryamba

Answer:

n=-7

Step-by-step explanation:

For this you can do the inverse of division which is division. So Divide -91 and 13. Which is -7.

7 0
3 years ago
Read 2 more answers
Can someone plz help me
dangina [55]

Answer:

4j+24k

Step-by-step explanation:

Expand by distributing terms.

4j+4\times 6k

4j+4×6k

2 Simplify  4\times 6k4×6k  to  24k24k.

4j+24k

4j+24k

3 0
3 years ago
Need to find what x and y equals
dexar [7]

The two angles with x in them are the same. Set them equal to each other and solve for x:

4x-5 = 3x + 11

Add 5 to both sides:

4x = 3x +16

Subtract 3x from both sides:

x = 16

Now solve for the angle next to y:

4x - 5 = 4(16) -5 = 64 -5 = 59

Y + 59 make a straight line and equals 180, so subtract 59 from 180 to get y:

y = 180 - 59 = 121

Answers:

X = 16, y = 121

8 0
3 years ago
Prove the identity secxcscx(tanx+cotx)=2+tan^2x+cot^2x
svetlana [45]
Hello,

sec(x)= \dfrac{1}{cos(x)} \\

cosec(x)= \dfrac{1}{sin(x)} \\

sec(x)*cosec(x)*(tg(x)+cotg(x))=\dfrac{1}{cos(x)}* \dfrac{1}{sin(x)}*( \frac{sin(x)}{cos(x)} +\frac{cos(x)}{sin(x)})\\

= \dfrac{sin^2(x)+cos^2(x)}{sin^2x*cos^2x} \\

= \dfrac{1}{sin^2x*cos^2x} \\


==============================================================
2+tg^2(x)+cotg^2(x)=2+ \dfrac{sin^2x}{cos^2x} + \dfrac{cos^2x}{sin^2x} \\

=2+ \dfrac{sin^4x+cos^4x}{sin^2x*cos^2x} \\

=\dfrac{2*sin^2x*cos^2x+sin^4x+cos^4x}{sin^2x*cos^2x} \\

= \dfrac{(sin^2x+cos^2x)^2}{sin^2x*cos^2x}} \\

= \dfrac{1}{sin^2x*cos^2x}} 

8 0
3 years ago
Read 2 more answers
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