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3 years ago

Find the result when 7x + 12 is subtracted from 11x – 10. - We have to use distributing

Mathematics

2 answers:

Sveta_85 [38]3 years ago

8 0

Answer:

$-4x+2$

Step-by-step explanation:

STEP 1: Subtract 11x from 7x.

STEP 2: Subtract 10 from 12.

Bingel [31]3 years ago

3 0

Answer:

-4x + 22

Step-by-step explanation:

7x + 12

- 11x - 10

-4x + 22

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Write 400 in scientific notation.

OlgaM077 [116]

= 4 × 10 to the power of 2

Step-by-step explanation:

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3 years ago

Sinx = 1/2, cosy = sqrt2/2, and angle x and angle y are both in the first quadrant.

Leviafan [203]

Answer:

Option D. 3.73

Step-by-step explanation:

we know that

$tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$

and

$sin^{2}(\alpha)+cos^{2}(\alpha)=1$

step 1

Find cos(X)

we have

$sin(x)=\frac{1}{2}$

we know that

$sin^{2}(x)+cos^{2}(x)=1$

substitute

$(\frac{1}{2})^{2}+cos^{2}(x)=1$

$cos^{2}(x)=1-\frac{1}{4}$

$cos^{2}(x)=\frac{3}{4}$

$cos(x)=\frac{\sqrt{3}}{2}$

step 2

Find tan(x)

$tan(x)=sin(x)/cos(x)$

substitute

$tan(x)=1/\sqrt{3}$

step 3

Find sin(y)

we have

$cos(y)=\frac{\sqrt{2}}{2}$

we know that

$sin^{2}(y)+cos^{2}(y)=1$

substitute

$sin^{2}(y)+(\frac{\sqrt{2}}{2})^{2}=1$

$sin^{2}(y)=1-\frac{2}{4}$

$sin^{2}(y)=\frac{2}{4}$

$sin(y)=\frac{\sqrt{2}}{2}$

step 4

Find tan(y)

$tan(y)=sin(y)/cos(y)$

substitute

$tan(y)=1$

step 5

Find tan(x+y)

$tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$

substitute

$tan(x+y)=[1/\sqrt{3}+1}]/[{1-1/\sqrt{3}}]=3.73$

7 0

3 years ago

Which is true of a circle?

MrRa [10]

Answer:

Any two points on the circumference of the circle are the same distance from the center of the circle.

Step-by-step explanation:

Any two points on the circumference of the circle are the same distance from the center of the circle.

5 0

3 years ago

Please help me oh please

mars1129 [50]

Answer: D or y= 4/3+ 18

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Write the equation of the circle with the given center and radius.

34kurt

Step-by-step explanation:

The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$

Center = (h, k)

Center = (0,2)

h = 0

k = 2

r = 5

$(x - h)^2 + (y - k)^2 = r^2$

$(x - 0)^2 + (y - 2)^2 = 5^2$

Center = (-4,-5)

h = -4

k = -5

r = $\sqrt{2}$

$(x - h)^2 + (y - k)^2 = r^2$

$(x - (-4))^2 + (y - (-5))^2 = \sqrt{2}^2$

$(x + 4))^2 + (y + 5))^2 = \sqrt{2}^2$

Center =(-1,3)

h = -1

k = 3

r = 8

$(x - h)^2 + (y - k)^2 = r^2$

$(x - (-1))^2 + (y - 3)^2 = 8^2$

$(x + 1)^2 + (y - 3)^2 = 8^2$

Center: (9,0)

h = 9

k = 0

r = $\sqrt{3}$

$(x - h)^2 + (y - k)^2 = r^2$

$(x - 9)^2 + (y - 0)^2 = \sqrt{3}^2$

4 0

3 years ago