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kow [346]
2 years ago
5

Reduce into lowest terms

Mathematics
1 answer:
Pavel [41]2 years ago
7 0

Answer:

4/6 = 2/3

10/12 = 5/6

18/24 = 3/4

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Factor completely −2x^3 + 6x^2. (1 point)
VashaNatasha [74]

Answer:

-2 x^2 (x - 3) thus the answer is c:)

Step-by-step explanation:

Factor the following:

6 x^2 - 2 x^3

Factor -2 x^2 out of 6 x^2 - 2 x^3:

Answer: -2 x^2 (x - 3)

5 0
2 years ago
Read 2 more answers
Helpppppppp pleaseeeee
lakkis [162]

Answer:

108 cm^2

Step-by-step explanation:

A= 1/2 b×h

so

1/2 12×18

3 0
3 years ago
=
sergij07 [2.7K]

Answer:

I. L = 17 cm.

II. W = 11 cm.

Step-by-step explanation:

Let the length of the rectangle be L.

Let the width of the rectangle be W.

Given the following data;

Perimeter of rectangle = 56cm

Translating the word problem into an algebraic expression, we have;

L = W + 6 .....equation 1

The perimeter of a rectangle is given by the formula;

P = 2(L + W)

56 = 2(L + W) ......equation 2

Substituting eqn 1 into eqn 2, we have;

56 = 2(W + 6 + W)

56 = 2(6 + 2W)

Opening the bracket, we have;

56 = 12 + 4W

4W = 56 - 12

4W = 44

W = 44/4

W = 11 cm

Next, we would find the length of the rectangle;

From eqn 1;

L = W + 6

L = 11 + 6

L = 17 cm

7 0
3 years ago
Solve (x + 1 < 4) ∩ (x - 8 > -7).
Monica [59]

1 < x < 3 because

( x < 3 ) intersecting ( x > 1)

3 0
2 years ago
Read 2 more answers
if sin theta=1/root 3, then find the value of (tan theta + 1/cos theta)^2 + (tan theta - 1/cos theta)^2
Makovka662 [10]

Answer:

4

Step-by-step explanation:

sin(\theta)=\frac{opposite}{hypotenuse}=\frac{1}{\sqrt{3} }\\  \\Therefore, opposite= 1, hypotenuse=\sqrt{3}. \ using \ Pythagoras:\\\\Hypotenuse^2=Adjacent^2+Opposite^2\\\\(\sqrt{3})^2 =Adjacent^2+1^2\\\\3=Adjacent^2+1\\\\Adjacent^2=3-1=2\\\\Adjacent=\sqrt{2}

cos(\theta)=\frac{adjacent}{hypotenuse}=\frac{\sqrt{2} }{\sqrt{3}}\\  \\tan(\theta)=\frac{opposite}{adjacent } = \frac{1}{\sqrt{2}}  \\\\Therefore:\\\\(tan(\theta)+\frac{1}{cos(\theta)})^2+ (tan(\theta)-\frac{1}{cos(\theta)})^2=( \frac{1}{\sqrt{2}}+\frac{1}{ \frac{\sqrt{2} }{\sqrt{3}}} )^2+( \frac{1}{\sqrt{2}}-\frac{1}{ \frac{\sqrt{2} }{\sqrt{3}}} )^2\\\\=( \frac{1}{\sqrt{2}}+ \frac{\sqrt{3} }{\sqrt{2}} )^2+( \frac{1}{\sqrt{2}}- \frac{\sqrt{3} }{\sqrt{2}} )^2

=(\frac{1+\sqrt{3} }{\sqrt{2} } )^2+(\frac{1-\sqrt{3} }{\sqrt{2} } )^2=\frac{1+2\sqrt{3}+3 }{2} +\frac{1-2\sqrt{3}+3 }{2} =\frac{1+1+2\sqrt{3}-2\sqrt{3}+3+3  }{2} =\frac{8}{2}=4

7 0
3 years ago
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