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3 years ago

5

I need help I give thanks and follow

2 answers:

olga_2 [115]3 years ago

5 0

Answer:

73,780

Step-by-step explanation:

2,108 x 35 = 73,780

Wewaii [24]3 years ago

5 0

Answer:

73,780

Step-by-step explanation:

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I need help with this any help will be appreciated

Mars2501 [29]

Are you trying to solve for x and y?

y=2x+1; x+2y=17
sub in one variable in the other equation.
x+2(2x+1)=17
x+4x+2=17
5x=15
x=3

now that you have x value, plug into the original equation to find y and check with the other.
y=2(3)+1 = 7
x+2y=17 = 3 + 2(7) = 17 ; 17=17

7 0

3 years ago

Not gonna sugar coat it <br> im cheeting...

Neporo4naja [7]

Answer:

The answer is <u><em>-0.44</em></u>.

Hope that helps. x

6 0

2 years ago

Read 2 more answers

-6 + 3a - b - 4a + 4b + 5

Naddika [18.5K]

Step-by-step explanation:

hope this may help you tq have a nice day

5 0

3 years ago

The mean length of 4 childrens' big finger is 14cm. The mean length of 9 adults' big finger is 16.1cm. What is the mean length (

Mamont248 [21]

Answer:

The mean length of the 13 people's big finger is 15.45 cm

Step-by-step explanation:

Given;

mean length of 4 childrens' big finger, x' = 14cm

mean length of 9 adults' big finger is 16.1cm, x'' = 16.1cm

Let the total length of the 4 childrens' big finger = t

$x' = \frac{t}{n} \\\\x' = \frac{t}{4}\\\\t = 4x'\\\\t = 4 *14\\\\t = 56 \ cm$

Let the total length of the 9 adults' big finger = T

$x'' = \frac{T}{N} \\\\x'' = \frac{T}{9}\\\\T = 9x''\\\\T = 9*16.1\\\\T = 144.9 \ cm$

The total length of the 13 people's big finger = t + T

= 56 + 144.9

=200.9 cm

The mean length of these 13 people's big finger;

x''' = (200.9) / 13

x''' = 15.4539 cm

x''' = 15.45 cm (2 DP)

Therefore, the mean length of the 13 people's big finger is 15.45 cm

3 0

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