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shepuryov [24]
2 years ago
6

Help me this is very hard

Mathematics
1 answer:
guapka [62]2 years ago
4 0

Answer:

B. 60

Step-by-step explanation:

1. take the square root of 64

2. multiply 8*10 and 5*4

3. subtract

Square root of 64 is 8 so,

10*8 - 5*4

80 - 20 =

60

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Indicate the equation of the given line in standard form. The line with slope and containing the midpoint of the segment whose e
Tems11 [23]

Answer:

x + y = -1

Step-by-step explanation:

To write the equation of a line, find the slope and use it with a point in the point slope form.

You can find the slope using the slope formula.

m = \frac{y_2-y_1}{x_2-x_1} = \frac{5--3}{-6-2} = \frac{8}{-8}=-1

Substitute m = -1 and the point (2,-3) into the point slope form.

y - y_1 = m(x-x_1)\\y --3 = -1(x-2)\\y+3=-1(x-2)

Convert to standard form by applying the distributive property and rearranging the terms.

y+3 = -1(x-2)

y + 3 = -x + 2

x + y + 3 = 2

x + y = -1

8 0
3 years ago
Please help me! yah thats all i just need help
Verdich [7]

Answer:

WT=63

Step-by-step explanation:

\frac{56}{5 \times x + 3}  =  \frac{40}{4 \times x - 3}  \\ x = 12

So

5 \times x + 3 = 5 \times 12 + 3 = 63

6 0
3 years ago
Really need help with this ​
IrinaVladis [17]

Answer:

36.89°

Step-by-step explanation:

tan x = 15/20=3/4

x = 36.89

5 0
2 years ago
What is 5\9 as a decimal
Eva8 [605]

Answer:

0. 5 the hyphen is over the 5 btw

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A drop of oil makes a circular ring in a water puddle. The radius of the oil ring expands at 1.5 cm per second. If r(t)=1.5t and
Readme [11.4K]

Answer:

a(t) =n(0.25t^2)

Step-by-step explanation:

Given

r(t) = 0.5t

a(r) = nr^2

Required

Determine the area in terms of time

From the question, we have:

r(t) = 0.5t -> radius per time

and

a(r) = nr^2 --> area from radius

The interpretation of the question is to find the composite function: a(t)

If a(r) = nr^2, and r(t) = 0.5t, the

Substitute 0.5t for r

a(t) =n(0.5t)^2

a(t) =n*0.25t^2

a(t) =n(0.25t^2)

5 0
3 years ago
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