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3 years ago

12

Can 7 over 63 be simplified

2 answers:

VMariaS [17]3 years ago

7 0

It is possible. It's 1/9. because 63/7=9. So 1/9 is the answer. When you get to these questions take the bigger number and divide it by the smaller one.

anzhelika [568]3 years ago

6 0

Yes 7 over 63 can be simplified

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Can someone please help me on this one problem?

UNO [17]

The angles of a quadrilateral sum to 360 degrees. A pentagon's exterior angles are 108 degrees. A hexagon's exterior angles are 60 degrees. So, <BAL AND <KLA are 60+108 or 168. 168 + 168 = 336. 360-336 = 24. If you write m<ABK as x, you can write the sum of <ABK and <LKB as 2x. If 2x=24, x=12. So, m<ABK is 12 degrees.

4 0

3 years ago

In a school, the ratio of musicians to athletes is 3 : 1. The number of musicians is how many times the number of athletes? ANS

loris [4]

Answer:

The number of musicians in the school is 3 times the number of athletes

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

3. Rectangle ABCD is shown on the coordinate grid below.

VMariaS [17]

Answer:

A. 8 units

Step-by-step explanation:

Coordinates of A = (-2, -2)

Coordinates of C = (5, 2)

Use distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ to calculate the length of diagonal $\overline{AC}$ .

Let,

$A(-2, -2) = (x_1, y_1)$

$C(5, 2) = (x_2, y_2)$

$d = \sqrt{(5 -(-2))^2 + (2 -(-2))^2}$

$d = \sqrt{(7)^2 + (4)^2}$

$d = \sqrt{49 + 16} = \sqrt{65}$

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Therefore, the closest to the length of diagonal $\overline{AC} = 8 units$

4 0

3 years ago

Tn = -2n + 3, find t4

yanalaym [24]

Substitute n = 4

t(4)=-2(4)+3

-8+3= -5

therefore t(4) = -5

4 0

3 years ago

Point B has coordinates (4,1). The x-coordinate of point A is -4. The distance between point A and point B is 10 units. What a

klasskru [66]

Given:

Point B has coordinates (4,1).

The x-coordinate of point A is -4.

The distance between point A and point B is 10 units.

To find:

The possible coordinates of point A.

Solution:

Let the y-coordinate of point A be y. Then the two points are A(-4,y) and B(4,1).

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The distance between point A and point B is 10 units.

$\sqrt{(4-(-4))^2+(1-y)^2}=10$

Taking square on both sides, we get

$(8)^2+(1-y)^2=100$

$(1-y)^2=100-64$

$(1-y)^2=36$

Taking square root on both sides, we get

$(1-y)=\pm \sqrt{36}$

$-y=\pm 6-1$

$y=1\mp 6$

$y=1-6$ and $y=1+6$

$y=-5$ and $y=7$

Therefore, the possible coordinates of point A are either (-4,-5) or (-4,7).

7 0

3 years ago