The drawing below shows the dimensions of skims kite what is the area of Kim's kite

Mathematics

1 answer:

Leno4ka [110]2 years ago

4 0

Answer:

48 square inches

Step-by-step explanation:

Given

See attachment for kite

Required

The area of the kite

The given kite has a triangular shape.

So, the area is:

$Area = 0.5 * base * height$

From the attachment

$base = 12in$

$height = 8in$

So:

$Area = 0.5 * 12in* 8in$

$Area = 48in^2$

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Find the roots for (3x + 6)(x - 4) = 0.

Elden [556K]

Answer:

Its 2, 3

Step-by-step explanation:

3 0

2 years ago

HELP PLEASE !!!!

dimulka [17.4K]

1. Let the shorter piece be x in. long
the longer piece will be x + 13 in. long
and the two pieces add up to be 47 in., so
x + ( x + 13 ) = 47
2x = 47 - 13
2x = 34
x = 17
the longer piece is x + 13 in. long,, or 17 + 13 = 30 in. long

2. number of dimes: x
number of quarters: y
x + y = 27....(1)
10x + 25y = 405...(2)
(1)*10: 10x + 10y = 270....(3)
(2) - (3):15y = 135
y = 9
plug y = 9 into (1), you will find that x = 18
so 18 dimes, 9 quarters

5 0

2 years ago

Read 2 more answers

Can someone help solve both of these? ASAP

zhuklara [117]

Question 4

$\sin 30^{\circ}=\frac{y}{14\sqrt{3}}\\\\\frac{1}{2}=\frac{y}{14\sqrt{3}}\\\\\boxed{y=7\sqrt{3}}\\\\\\\\\cos 30^{\circ}=\frac{x}{14\sqrt{3}}\\\\\frac{\sqrt{3}}{2}=\frac{x}{14\sqrt{3}}\\\\\boxed{x=21}$

Question 5

$\cos 45^{\circ}=\frac{x}{28}\\\\\frac{1}{\sqrt{2}}=\frac{x}{28}\\\\x=\frac{28}{\sqrt{2}}\\\\\boxed{x=14\sqrt{2}}\\\\\\\\\sin 45^{\circ}=\frac{y}{28}\\\\\frac{1}{\sqrt{2}}=\frac{y}{28}\\\\y=\frac{28}{\sqrt{2}}\\\\\boxed{y=14\sqrt{2}}$