Ask question

Ask question

All categories

3 years ago

12

a checheers board is 8 square long and 7 squares wide. The area of each square is 14 square centimeter. Estimate the perimeter o

f the checkers board to the nearest tenth of a centimeter

1 answer:

Lesechka [4]3 years ago

4 0

Turn 14 into 16, 4 x 4 = 16, 4 x 8 = 32, 32 x 4 = 128cm

You might be interested in

bernite is selling candy bars for a school fundraiser. the school paid $20 for a box of 15 king size candy bars and bernite sell

NikAS [45]

The school originally used $20 for the purchase of the candy.
Use the formula Y=2x to find how much money is made (y) from the number of candy bars (x)
Sub in the 15 bars for t
Y=2(15)
Y=30
Therefore the school made $10, because they earned an extra $10 in comparison to the amount they bought the candy bars for, which was $20.
Hope this helps!

6 0

3 years ago

A rectangle has base length

vekshin1

I believe that the answer is 22x

3 0

3 years ago

Two sides of a triangle are 4 cm and 7 cm. What can be the length of its third side to make the triangle possible?

Artyom0805 [142]

Answer:

4cm or 7 cm

Step-by-step explanation:

it can make 2 sides equal and make a isosceles triangle

8 0

3 years ago

Shaun plotted a point on the number line by drawing 5 equally spaced marks between 0 and 1 and placing a point in the third mark

SIZIF [17.4K]

If it is 5 equally spaced marks, then each mark represents 1/6

0 -- 1/6 -- 2/6 -- 3/6 -- 4/6 -- 5/6 -- 1

so the third mark would represent 3/6 which reduces to 1/2

4 0

3 years ago

Use the diagram to complete the statement. Triangle J K L is shown. Angle K J L is a right angle. Angle J K L is 52 degrees and

zzz [600]

Answer:

$\bold{sin(38^\circ)=cos(52^\circ)}$

Step-by-step explanation:

Given that $\triangle KJL$ is a right angled triangle.

$\angle JKL = 52^\circ\\\angle KLJ = 38^\circ$

and

$\angle KJL = 90^\circ$

Kindly refer to the attached image of $\triangle KJL$ in which all the given angles are shown.

To find:

sin(38°) = ?

a) cos(38°)

b) cos(52°)

c) tan(38°)

d) tan(52°)

Solution:

Let us use the trigonometric identities in the given $\triangle KJL$ .

We have to find the value of sin(38°).

We know that sine trigonometric identity is given as:

$sin\theta =\dfrac{Perpendicular}{Hypotenuse}$

$sin(\angle JLK) = \dfrac{JK}{KL}\\OR\\sin(38^\circ) = \dfrac{JK}{KL}$ ....... (1)

Now, let us find out the values of trigonometric functions given in options one by one:

$cos\theta =\dfrac{Base}{Hypotenuse}$

$cos(\angle JLK) = \dfrac{JL}{KL}\\OR\\cos(38^\circ) = \dfrac{JL}{KL}$ ....... (2)

By (1) and (2):

sin(38°) $\neq$ cos(38°).

$cos(\angle JKL) = \dfrac{JK}{KL}\\OR\\cos(52^\circ) = \dfrac{JK}{KL}$ ...... (3)

Comparing equations (1) and (3):

we get the both are same.

$\therefore \bold{sin(38^\circ)=cos(52^\circ)}$

6 0

3 years ago