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3 years ago

Due tomorrow deeply appreciated if done

Mathematics

1 answer:

fiasKO [112]3 years ago

6 0

This is unreadable try submitting a more clear picture. i can bairly read the first problem i think it says 8(3x+4)=2(17x????)

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Ricky caught a snake that was 58.2 centimeters long. Which of these lengths has an 8 with a value of 1/10 the value of the 8 in

Anettt [7]

In every number, set every digit to zero that is not 8. Then look at the question again.

"Which of these lengths has 1/10 the value of 8.0?"

A. 800
B. 0.08
C. 80
D. 0.80

We hope the answer D is obvious to you.

... 0.80 = (1/10)×8.0

___

The appropriate choice is ...

... D. 209.86

8 0

3 years ago

What is 1/3 multiply by 5

Irina-Kira [14]

Answer:

$\frac{5}{3}$

hope that helps uhh :)

3 0

2 years ago

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In Mrs.Hu's classroom, 4/5 of the students have a dog as a pet. Of the students who have a dog as a pet also have cat as a pet.

Travka [436]

Answer:

This problem is incomplete, we do not know the fraction of the students that have a dog and also have a cat. Suppose we write the problem as:

"In Mrs.Hu's classroom, 4/5 of the students have a dog as a pet. X of the students who have a dog as a pet also have cat as a pet. If there are 45 students in her class, how many have both a dog and a cat as pets?"

Where X must be a positive number smaller than one, now we can solve it:

we know that in the class we have 45 students, and 4/5 of those students have dogs, so the number of students that have a dog as a pet is:

N = 45*(4/5) = 36

And we know that X of those 36 students also have a cat, so the number of students that have a dog and a cat is:

M = 36*X

now, we do not have, suppose that the value of X is 1/2 ("1/2 of the students who have a dog also have a cat")

M = 36*(1/2) = 18

So you can replace the value of X in the equation and find the number of students that have a dog and a cat as pets.

5 0

2 years ago

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An angle bisector of a triangle divides the opposite side of the triangle into segments 5 cm and 3 cm long. A second side of the

attashe74 [19]

There is a little-known theorem to solve this problem.

The theorem says that
In a triangle, the angle bisector cuts the opposite side into two segments in the ratio of the respective sides lengths.

See the attached triangles for cases 1 and 2. Let x be the length of the third side.

Case 1:
Segment 5cm is adjacent to the 7.6cm side, then
x/7.6=3/5 => x=7.6*3/5=4.56 cm

Case 2:
Segment 3cm is adjacent to the 7.6 cm side, then
x/7.6=5/3 => x=7.6*5/3=12.67 cm

The theorem can be proved by considering the sine rule on the adjacent triangles ADC and BDC with the common side CD and equal angles ACD and DCB.

6 0

2 years ago

Find the area of the shape shown below.

vlabodo [156]

Answer:

6 $cm^{2}$ (assuming the units is in centimetres)

Step-by-step explanation:

Hi, hope this helps!

Method 1:

Calculate this shape as a trapezium. This is a trapezium because it has one pair of parallel sides. The way to calculate the area of a trapezium is this:

- Half the sum of the parallel sides

- Multiply the perpendicular height between them

In this case, 4 and 2 are parallel, so we add them together (4 + 2 = 6) then we divide by two (6 ÷ 2 = 3). Then we multiply our answer by the perpendicular length between the two parallel sides (assuming the side on the left is at a right-angle, 3 x 2 = 6).

Method 2:

Calculate this shape as a rectangle + a triangle. If you split the shape so the pointy bit on the right becomes a triangle and the left becomes a square, you can calculate the area without knowing the formula for calculating a trapezium (see above ^).

- Area of the square / rectangle = length x width = 2 x 2 = 4

- Area of the triangle = 1/2 x length x width = 2 x 2 x 1/2 = 4 x 1/2 = 4 ÷ 2 = 2

Then we add the two areas together (4 + 2 = 6).

I hope this helped and I wasn't waffling on :)

Bluey

8 0

3 years ago

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