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3 years ago

9.35 as a fraction in simpliest form

2 answers:

8 0

We know that
the number $9.35$
is equal to
$9+0.35$
and
$0.35$ is equal to
$35/100$
divide by 5 both members
$\frac{35}{100} = \frac{7}{20}$

therefore

the number $9.35$ is equal to $9 \frac{7}{20}$

Flura [38]3 years ago

7 0

The correct answer is:

9 7/20.

Explanation:

Reading this decimal, we would say "nine and thirty-five hundredths." This means we can write it as a fraction in the form

9 35/100.

To simplify the fraction, we look for common factors in the numerator and denominator. The numerator ends in 5 and the denominator ends in 0, so both are divisible by 5:

(35÷5)/(100÷5) = 7/20

From this point, the only numbers that are factors of 7 are 1 and 7; while 1 is a factor of everything, 7 will not go into 20, so this does not simplify further.

Thus we have 9 7/20.

You might be interested in

Find the solution in the set W. 3 + x = x + 3

Keith_Richards [23]

Your answer will be 0.
Because if 3+x is the same as x+3. That is literally the same as 3+0=3. You can plug the 0 into the variable

4 0

3 years ago

I neeeeed help please

kodGreya [7K]

9514 1404 393

Answer:

11) C: (a, 4), (b, 3), (c, 2), (d, 1); V: (a, c), (b, d), (4, 2), (3, 1);

AI: (c, 4), (d, 3); AE: (a, 2), (b, 1); SI: (c, 3), (d, 4); SE: (a, 1), (b, 2)

12) : C: (a, y), (b, x), (c, w), (d, z); V:(a, c), (b, d), (w, y), (x, z);

AI: (b, z), (c, y); AE: (a, w), (d, x); SI: (b, y), (c, z); SE: (a, x), (d, w)

Step-by-step explanation:

This is a vocabulary question. It is intended to see if you understand the meaning of the names given to the different angle pairs in this geometry.

All of the pairs of angles filling the first 4 blanks (corresponding, vertical, alt. int., alt. ext.) are congruent pairs of angles. The pairs of angles filling the last two blanks (same-side ...) are supplementary angles (total 180°).

Corresponding angles lie in the same direction from the point of intersection. For example, the upper-left angles are corresponding.

Vertical angles are formed from opposite rays.

Alternate refers to angles on opposite sides of the transversal. (Same-side refers to angles on the same side of the transversal. Same-side angles are also called consecutive angles.)

Interior refers to angles that are between the parallel lines. Exterior refers to angles outside the parallel lines.

The figures are a bit fuzzy. We assume the angle designations are (CW around the intersection point from upper left, left group first) ...

Figure 11) {a, b, c, d}, {4, 3, 2, 1}

Figure 12) {d, c, b, a}, {z, w, x, y}

If this is not correct, you will need to make appropriate substitutions in the pairs given below.

11) Corresponding: (a, 4), (b, 3), (c, 2), (d, 1)

Vertical: (a, c), (b, d), (4, 2), (3, 1)

Alternate Interior: (c, 4), (d, 3)

Alternate Exterior: (a, 2), (b, 1)

Same-side Interior: (c, 3), (d, 4)

Same-side Exterior: (a, 1), (b, 2)

12) Corresponding: (a, y), (b, x), (c, w), (d, z)

Vertical: (a, c), (b, d), (w, y), (x, z)

Alternate Interior: (b, z), (c, y)

Alternate Exterior: (a, w), (d, x)

Same-side Interior: (b, y), (c, z)

Same-side Exterior: (a, x), (d, w)

_____

Additional comment

The vocabulary is used to cite various theorems supporting claims regarding angle relationships. In practice, when the lines are parallel, all obtuse angles related to a transversal are congruent, all acute angles are congruent, and the acute angles are supplementary to the obtuse angles. (If the angles are 90°, neither acute nor obtuse, then they are all congruent.)

3 0

3 years ago

ASAP 30 + Brainliest Please only solve 2 - 5

hichkok12 [17]

QUESTION 2a

We want to find the area of the given right angle triangle.

We use the formula

$Area=\frac{1}{2}\times base\times height$

The height of the triangle is $=a cm$ .

The base is $12cm$ .

We substitute the given values to obtain,

$Area=\frac{1}{2}\times 12\times a cm^2$ .

This simplifies to get an expression for the area to be

$Area=6a cm^2$ .

QUESTION 2b

The given diagram is a rectangle.

The area of a rectangle is given by the formula

$Area=length \times width$

The length of the rectangle is $l=7cm$ and the width of the rectangle is $w=ycm$ .

We substitute the values to obtain the area to be

$Area=7 \times y$

The expression for the area is

$Area=7y$

QUESTION 2c.

The given diagram is a rectangle.

The area of a rectangle is given by the formula

$Area=length \times width$

The length of the rectangle is $l=2x cm$ and the width of the rectangle is $w=4 cm$ .

We substitute the values to obtain the area to be

$Area=2x \times 4$

The expression for the area is

$Area=8x$

QUESTION 2d

The given diagram is a square.

The area of a square is given by,

$Area=l^2$ .

where $l=b m$ is the length of one side.

The expression for the area is

$Area=b^2 m^2$

QUESTION 2e

The given diagram is an isosceles triangle.

The area of this triangle can be found using the formula,

$Area=\frac{1}{2}\times base\times height$ .

The height of the triangle is $4cm$ .

The base of the triangle is $6a cm$ .

The expression for the area is

$Area=\frac{1}{2}\times 6a \times 4cm^2$

$Area=12a cm^2$

QUESTION 3a

Perimeter is the distance around the figure.

Let P be the perimeter, then

$P=x+x+x+x$

The expression for the perimeter is

$P=4x mm$

QUESTION 3b

The given figure is a rectangle.

Let P, be the perimeter of the given figure.

$P=L+B+L+B$

This simplifies to

$P=2L+2B$

$P=2(L+B)$

QUESTION 3c

The given figure is a parallelogram.

Perimeter is the distance around the parallelogram

$Perimeter=3q+P+3q+P$

This simplifies to,

$Perimeter=6q+2P$

$Perimeter=2(3q+P)$

QUESTION 3d

The given figure is a rhombus.

The perimeter is the distance around the whole figure.

Let P be the perimeter. Then

$P=5b+5b+5b+5b$

This simplifies to,

$P=20b mm$

QUESTION 3e

The given figure is an equilateral triangle.

The perimeter is the distance around this triangle.

Let P be the perimeter, then,

$P=2x+2x+2x$

We simplify to get,

$P=6x mm$

QUESTION 3f

The figure is an isosceles triangle so two sides are equal.

We add all the distance around the triangle to find the perimeter.

This implies that,

$Perimeter=3m+5m+5m$

$Perimeter=13m mm$

QUESTION 3g

The given figure is a scalene triangle.

The perimeter is the distance around the given triangle.

Let P be the perimeter. Then

$P=(3x+1)+(2x-1)+(4x+5)$

This simplifies to give us,

$P=3x+2x+4x+5-1+1$

$P=9x+5$

QUESTION 3h

The given figure is a trapezium.

The perimeter is the distance around the whole trapezium.

Let P be the perimeter.

Then,

$P=m+(n-1)+(2m-3)+(n+3)$

We group like terms to get,

$P=m+2m+n+n-3+3-1$

We simplify to get,

$P=3m+2n-1$ mm

QUESTION 3i

The figure is an isosceles triangle.

We add all the distance around the figure to obtain the perimeter.

Let $P$ be the perimeter.

Then $P=(2a-b)+(a+2b)+(a+2b)$

We regroup the terms to get,

$P=2a+a+a-b+2b+2b$

This will simplify to give us the expression for the perimeter to be

$P=4a+3b$ mm.

QUESTION 4a

The given figure is a square.

The area of a square is given by the formula;

$Area=l^2$

where $l=2m$ is the length of one side of the square.

We substitute this value to obtain;

$Area=(2m)^2$

This simplifies to give the expression of the area to be,

$Area=4m^2$

QUESTION 4b

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$ .

where $l=5a cm$ is the length of the rectangle and $w=6cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =5a \times 6$

$Area =30a cm^2$

QUESTION 4c

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$ .

where $l=7y cm$ is the length of the rectangle and $w=2x cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =7y \times 2x$

The expression for the area is

$Area =14xy cm^2$

QUESTION 4d

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$ .

where $l=3p cm$ is the length of the rectangle and $w=p cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =3p \times p$

The expression for the area is

$Area =3p^2 cm^2$

See attachment for the continuation

6 0

3 years ago

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marta [7]

Answer:

Step-by-step explanation:12

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1 ? I checked on the calculator and it said 1

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3 years ago