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2 years ago

Help ! 7p^2-38p-24=0

2 answers:

8 0

7p^2-38p-24=0

All you need is to factor it, as follows

(7p + 4) (p - 6) = 0

Use FOIL to make sure that you got the correct factoring

Hope that helps

ivolga24 [154]2 years ago

6 0

What is this. what are post to do.

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ANSWER 3-8 FOR BRAINLIESET ANSWER AND ALL POINTS!!!! ALL INCOMPLETE/WRONG ANSWERS WILL BE REPORTED URGENT HELP REQUIRED.

Verdich [7]

3. The correct answer is option A (10 in)

4. The correct answer is option C (5 in)

5. Cannot see this question.

6. The correct answer is option B (17.5 ft)

7. The correct answer is option A (5)

8. The correct answer is option A (3.375)

Hope this helped. Have a great day!

4 0

3 years ago

ABC is an isosceles AB= 3x-4 and BC= 5x-10 what is AB?

kakasveta [241]

3x-4=5x-10
Subtract 3x from both sides
-4=2x-10
Add 10 to both sides
6= 2x
X= 3
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3 0

2 years ago

Read 2 more answers

Suppose that S is the set of successful students in a classroom, and that F stands for the set of freshmen students in that clas

lora16 [44]

Answer:

b) 24

Step-by-step explanation:

We solve building the Venn's diagram of these sets.

We have that n(S) is the number of succesful students in a classroom.

n(F) is the number of freshmen student in that classroom.

We have that:

$n(S) = n(s) + n(S \cap F)$

In which n(s) are those who are succeful but not freshmen and $n(S \cap F)$ are those who are succesful and freshmen.

By the same logic, we also have that:

$n(F) = n(f) + n(S \cap F)$

The union is:

$n(S \cup F) = n(s) + n(f) + n(S \cap F)$

In which

$n(S \cup F) = 58$

$n(s) = n(S) - n(S \cap F) = 54 - n(S \cap F)$

$n(f) = n(F) - n(S \cap F) = 28 - n(S \cap F)$

$n(S \cup F) = n(s) + n(f) + n(S \cap F)$

$58 = 54 - n(S \cap F) + 28 - n(S \cap F) + n(S \cap F)$

$n(S \cap F) = 24$

So the correct answer is:

b) 24

3 0

3 years ago

Two competitive neighbours build rectangular pools that cover the same area but are different shapes. Pool A has a width of (x +

GenaCL600 [577]

Answer:

a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

b) Area of pool A is equal to area of pool B equal to 24.44 meters.

Solution:

Let’s first calculate area of pool A .

Given that width of the pool A = (x+3)

Length of the pool A is 3 meter longer than its width.

So length of pool A = (x+3) + 3 =(x + 6)

Area of rectangle = length x width

So area of pool A =(x+6) (x+3) ------(1)

Let’s calculate area of pool B

Given that length of pool B is double of width of pool A.

So length of pool B = 2(x+3) =(2x + 6) m

Width of pool B is 4 meter shorter than its length,

So width of pool B = (2x +6 ) – 4 = 2x + 2

Area of rectangle = length x width

So area of pool B =(2x+6)(2x+2) ------(2)

Since area of pool A is equal to area of pool B, so from equation (1) and (2)

(x+6) (x+3) =(2x+6) (2x+2)

On solving above equation for x

(x+6) (x+3) =2(x+3) (2x+2)

x+6 = 4x + 4

x-4x = 4 – 6

x = $\frac{2}{3}$

Dimension of pool A

Length = x+6 = ( $\frac{2}{3}$ ) +6 = 6.667m

Width = x +3 = ( $\frac{2}{3}$ ) +3 = 3.667m

Dimension of pool B

Length = 2x +6 = 2( $\frac{2}{3}$ ) + 6 = $\frac{22}{3}$ = 7.333m

Width = 2x + 2 = 2( $\frac{2}{3}$ ) + 2 = $\frac{10}{3}$ = 3.333m

Verifying the area:

Area of pool A = ( $\frac{20}{3}$ ) x ( $\frac{11}{3}$ ) = $\frac{220}{9}$ = 24.44 meter

Area of pool B = ( $\frac{22}{3}$ ) x ( $\frac{10}{3}$ ) = $\frac{220}{9}$ = 24.44 meter

Summarizing the results:

(a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

(b)Area of pool A is equal to Area of pool B equal to 24.44 meters.

5 0

3 years ago

If you have two boxes,and in each box is 20 plates, how many boxes do you have

Free_Kalibri [48]

40 just multiply 20 by 2

4 0

3 years ago