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2 years ago

11

A test has twenty questions worth 100 points. The test consists of questions worth 3 points each and multiple choice questions w

orth 11 points each. How many multiple choice questions are on the test?

1 answer:

tigry1 [53]2 years ago

5 0

Answer:

depends how many questions are not multiple choice questions if i get that info then i can tell you how many are multiple questoions

Step-by-step explanation:

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How does decreasing only the mass change an objects density

Sladkaya [172]

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3 0

3 years ago

Do not know how to solve

djverab [1.8K]

Hehehe I wish A lallaalalalal

4 0

3 years ago

-0.111c + 1 = 0.889c

Elena-2011 [213]

Answer:

-0.111c + 1 = 0.889c = c=1

Subtract 0.889c from both sides:

−0.111c+1−0.889c=0.889c−0.889c

−c+1=0

Subtract 1 from both sides.:

−c+1−1=0−1

−c=−1

Divide both sides by -1.:

-c/1 = -1/-1

Therefore, your answer is c=1

7 0

3 years ago

3! + 4! = 5040 30 or 16

makkiz [27]

Answer:

None of those

Step-by-step explanation:

3! = 3*2*1 = 6

4! = 4*3*2*1 = 24

24 * 6 = 144

144 is none of the choices

6 0

3 years ago

Read 2 more answers

A rectangle has a length 6 more than it's width if the width is decreased by 2 and the length decreased by 4 the resulting has a

Rashid [163]

Answer:

Length of original rectangle: 11 units.

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

$\text{Perimeter of new rectangle}=20$

Step-by-step explanation:

Let x represent width of the original rectangle.

We have been given that a rectangle has a length 6 more than it's width. S the length of the original rectangle would be $x+6$ .

We have been given that when the width is decreased by 2 and the length decreased by 4 the resulting has an area of 21 square units.

The width of new rectangle would be $x-2$ .

The length of new rectangle would be $x+6-4=x+2$ .

The area of new rectangle would be $(x+2)(x-2)$ .

Now we will equate area of new rectangle with 21 and solve for x as:

$(x+2)(x-2)=21$

Applying difference of squares, we will get:

$x^2-2^2=21$

$x^2-4=21$

$x^2-4+4=21+4$

$x^2=25$

Since width cannot be negative, so we will take positive square root of both sides.

$\sqrt{x^2}=\sqrt{25}$

$x=5$

Therefore, the width of original rectangle is 5 units.

Length of the original rectangle would be $x+6\Rightarrow x+5=11$ .

Therefore, the length of original rectangle is 11 units.

$\text{Area of original rectangle}=5\times 11$

$\text{Area of original rectangle}=55$

Therefore, area of the original rectangle is 55 square units.

Now we will find ratio of the original rectangle area to the new rectangle area as:

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

We know that perimeter of rectangle is two times the sum of length and width.

$\text{Perimeter of new rectangle}=2((x+2)+(x-2))$

$\text{Perimeter of new rectangle}=2((5+2)+(5-2))$

$\text{Perimeter of new rectangle}=2(7+3)$

$\text{Perimeter of new rectangle}=2(10)$

$\text{Perimeter of new rectangle}=20$

Therefore, the perimeter of the new rectangle is 20 units.

7 0

3 years ago