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Nataliya [291]
2 years ago
7

PLEASE OLEASE HELP!!!

Mathematics
1 answer:
Andru [333]2 years ago
8 0

Answer:

answer = 7.5

Step-by-step explanation:

PG = 2 PR ( by Parallelogram law)

4x - 5 = 2 (x + 5)

4x - 5 = 2x + 10

4x - 2x = 10 + 5

2x = 15

x = 15/2

x = 7.5

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Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
SIZIF [17.4K]
C = 4M + 5
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answer is D. Last one.
6 0
3 years ago
An orchard has 651 orange trees. The number of rows exceeds the number of trees per row by 10. How many trees are there in a row
natima [27]

Answer:

The number of trees in each row is  21.

Step-by-step explanation:

Let us assume the number of trees in each row  = m

So, the number of rows in the orchard = Number of trees in each row + 10

or, total number of trees in each row = (10 +m)

Now, Total number of Trees = Number of Rows x Number of trees in 1 row

⇒ 651  = m (m + 10)

651 = m^2 + 10m\\\implies  m^2 + 10m - 651 =0\\or,   m^2 + 31m  21m - 651 =0\\or, m(m +31) -21(m+31) =0\\\implies (m +31) (m-21) = 0

⇒ (m +31) = 0, or (m-21) = 0

⇒ m= -31, or m = 21

Now, as m = The number of trees in each row, so m ≠ -31

So, m =  21

Hence, the number of trees in each row is m = 21.

8 0
3 years ago
The number of gallons of paint needed to cover a wall varies directly with the area of the wall. The Robertsons find that they u
Licemer1 [7]

Answer:

If 1/2 gallon covers 540 square feet, then 1 gallon will cover 1080 square feet.

Step-by-step explanation:

5 0
3 years ago
Find the area under the standard normal probability distribution between the following pairs of​ z-scores. a. z=0 and z=3.00 e.
prohojiy [21]

Answer:

a. P(0 < z < 3.00) =  0.4987

b. P(0 < z < 1.00) =  0.3414

c. P(0 < z < 2.00) = 0.4773

d. P(0 < z < 0.79) = 0.2852

e. P(-3.00 < z < 0) = 0.4987

f. P(-1.00 < z < 0) = 0.3414

g. P(-1.58 < z < 0) = 0.4429

h. P(-0.79 < z < 0) = 0.2852

Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

d.  z=0 and z=0.79

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

Thus;

P(0 < z < 0.79) = 0.7852- 0.5

P(0 < z < 0.79) = 0.2852

e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

f. z=−1.00 and z=0

From the standard normal distribution tables,

P(Z< -1.00) = 0.1587  and P(Z< 0) = 0.5

Thus;

P(-1.00 < z < 0 ) = 0.5 -  0.1586

P(-1.00 < z < 0) = 0.3414

g. z=−1.58 and z=0

From the standard normal distribution tables,

P(Z< -1.58) = 0.0571  and P(Z< 0) = 0.5

Thus;

P(-1.58 < z < 0 ) = 0.5 -  0.0571

P(-1.58 < z < 0) = 0.4429

h. z=−0.79 and z=0

From the standard normal distribution tables,

P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

P(-0.79 < z < 0) = 0.2852

8 0
3 years ago
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