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3 years ago

PLEASE help Asap please

Mathematics

1 answer:

puteri [66]3 years ago

8 0

I believe the answer would be 5 pages.
If you would like to see my work please reply saying so.
Hope this could help.

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5 barrels catch rainwater in the schoolyard. Four barrels are the same size, and the fifth barrel holds 10 liters of water. Comb

ch4aika [34]

Answer:

The volume each of the four barrels of the same size will be at least 47.5 liters.

Step-by-step explanation:

Let the volume of each of the four barrels is v liters.

Now, the amount of water that can be held in the fifth barrel is 10 liters. The total capacity of holding water for the 5 barrels combines is at least 200 liters.

So, from the above-given information, we can write the inequality to determine the volume of each of the four barrels will be

10 + 4v ≥ 200

Solving this we get,

4v ≥ 190

⇒ v ≥ 47.5

Therefore, the volume each of the four barrels of the same size will be at least 47.5 liters. (Answer)

3 0

2 years ago

Help me plz thanksss

natulia [17]

Answer:

0.5j+19

Step-by-step explanation:

6 0

2 years ago

15= x over 6 - 1

Diano4ka-milaya [45]

15 + 1 = x/6
16 = x/6
16 * 6 = x
x = 96

Answer: A)

3 0

3 years ago

Read 2 more answers

Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)

strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex

if $a>0$ -------> the parabola open upward (vertex is a minimum)

if $a$ -------> the parabola open downward (vertex is a maximun)

case A) $f(x)=(x-2)^{2}$

This is a vertical parabola open upward

the vertex is the point $(2,0)$

therefore

The function $f(x)=(x-2)^{2}$ does not have a vertex on the y-axis

case B) $f(x)=x(x+2)$

$f(x)=x(x+2)=x^{2}+2x$

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+1=x^{2}+2x+1$

Rewrite as perfect squares

$f(x)+1=(x+1)^{2}$

$f(x)=(x+1)^{2}-1$

the vertex is the point $(-1,-1)$

therefore

The function $f(x)=x(x+2)$ does not have a vertex on the y-axis

case C) $f(x)=(x-2)(x+2)$

$f(x)=(x-2)(x+2)=x^{2}-2^{2}$

$f(x)=x^{2}-4$

the vertex is the point $(0,-4)$

The x-coordinate of the vertex is equal to zero

therefore

The function $f(x)=(x-2)(x+2)$ has a vertex on the y-axis

case D) $f(x)=(x+1)(x-2)$

$f(x)=(x+1)(x-2)\\ \\f(x)= x^{2}-2x+x-2 \\ \\f(x)= x^{2} -x-2$

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+2= x^{2} -x$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+2+0.25= x^{2} -x+0.25$

$f(x)+2.25= x^{2} -x+0.25$

Rewrite as perfect squares

$f(x)+2.25= (x-0.50)^{2}$

$f(x)=(x-0.50)^{2}-2.25$

the vertex is the point $(0.5,-2.25)$

therefore

The function $f(x)=(x+1)(x-2)$ does not have a vertex on the y-axis

the answer is

$f(x)=(x-2)(x+2)$

6 0

3 years ago

Read 2 more answers

I really need help with this problem steps

melamori03 [73]

Solve for y by setting one equation =to X after that you can substitute the equation into the other one let's set the first one in X= FORM
X-7y=10
-2x+14y=-20

Add 7y to both sides of the first equation to let x stand alone
X=7y+10
Now you can substitute x on the second equation
-2 (7y+10)+14y=-20
Distribute
-14y-20+14y=-20
Add and simplify
Us cancel out =0
-20=-20
0=0
They are the same equations you can also divide the second equation by -2 which would make it look like this
X-7y=10
-2 (x-7y=10)
Let me know if I have answered your question

5 0

3 years ago