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allsm [11]
3 years ago
6

Select the correct answer from each drop down menu. The GCF of the numbers in the expression (32+16) is( ) The numbers left insi

de the parentheses after factoring out the GCF are ( ) and ( ) ​
Mathematics
1 answer:
lawyer [7]3 years ago
8 0

Answer:

Eu não entendo inglês!!

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5+b=25<br> b=20<br> 5+20=25<br> =?
Arisa [49]
The numbers are equal. 25=25
4 0
3 years ago
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4^x+7 = 8^2x-3 Solve each equation showing work please
RUDIKE [14]

If 4=2^2

8=2^3

(2^)2x+10=(2^)6x

(2^)6x-(2^)2x=10

(2^)2x×7=10, where x has no real value

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3 years ago
The ratio of a circle's radius to its diameter is always 1 to 2. In math class, Russell uses his compass to draw a circle with a
Semenov [28]

12 bc 6/12 equals 1/2

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3 years ago
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...........................
cupoosta [38]

Answer:

3/5 or 0.6

Step-by-step explanation:

Here, given the value of tan theta , we want to find the value of sine theta

Mathematically;

tan theta = 0pposite/adjacent

Sine theta = opposite/hypotenuse

Firstly we need the length of the hypotenuse

This can be obtained using the Pythagoras’ theorem which states that the square of the hypotenuse equals sum of the squares of the two other sides.

Let’s call the hypotenuse h

h^2 = 3^2 + 4^2

h^2 = 9 + 16

h^2 = 25

h = √(25)

h = 5

Now from the tan theta, we know that the opposite is 3

Thus, the value of the sine theta = 3/5 or simply 0.6

5 0
3 years ago
Solve for y 16y^2-25=0
Pepsi [2]

Answer:

\large\boxed{x=-\dfrac{5}{4}\ \vee\ x=\dfrac{5}{4}}

Step-by-step explanation:

16y^2-25=0\\\\METHOD\ 1:\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\16=4^2\ \text{and}\ 25=5^2\ \text{therefore we have}\\\\4^2y^2-5^2=0\\\\(4y)^2-5^2=0\\\\(4y-5)(4y+5)+0\iff4y-5=0\ \vee\ 4y+5=0\\\\4y-5=0\qquad\text{add 5 to both sides}\\4y=5\qquad\text{divide both sides by 4}\\\boxed{y=\dfrac{5}{4}}\\\\4y+5=0\qquad\text{subtract 5 from both sides}\\4y=-5\qquad\text{divide both sides by 4}\\\boxed{x=-\dfrac{5}{4}}

METHOD\ 2:\\\\16y^2-25=0\qquad\text{add 25 to both sides}\\\\16y^2=25\qquad\text{divide both sides by 16}\\\\y^2=\dfrac{25}{16}\to y=\pm\sqrt{\dfrac{25}{26}}\\\\y=-\dfrac{\sqrt{25}}{\sqrt{16}}\ \vee\ x=\dfrac{\sqrt{25}}{\sqrt{16}}\\\\\boxed{y=-\dfrac{5}{4}\ \vee\ x=\dfrac{5}{4}}

7 0
3 years ago
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