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3 years ago

7

Find the area of the figure shown. Please help!!!

1 answer:

LUCKY_DIMON [66]3 years ago

7 0

Answer:

See attached image!

Step-by-step explanation:

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A pair of jeans originally cost $80, but is on sale for 30% off. there is a coupon available for an additional 10% off of the sa

Lady_Fox [76]

30% = 0.3.

0.3*80 = 24, so we know that 30% of 80 is 24.

80 - 24 = 56, so after the sale, the price for a pair of jeans is $56.

10% = 0.1

0.1*56 = 5.6, so with the coupon, we get $5.60 off.

56 - 5.60 = 50.40

The jeans cost $50.40 after the sale and the coupon.

5 0

3 years ago

Help! This is due today and I'm stuck on this problem.

Varvara68 [4.7K]

Answer:

280.6 $m^2$

Step-by-step explanation:

<u>find the length of each side:</u>

you're only given h = 9 m

the formula for h is $h=\frac{\sqrt{3} }{2} *a$ , where a is a side length of the hexagon

plug h in:

$9=\frac{\sqrt{3}}{2}*a\\\\a=9*\frac{2}{\sqrt{3}}\\\\$

a ≈ 10.392 m

<u>now find the area of a triangle:</u>

1/2 (10.392)(9) = 46.764 $m^2$

multiply that by 6 and you get 280.584 ≈ 280.6 $m^2$

you can also use the regular hexagon area formula instead of finding the individual area and multiplying by 6: $\frac{3\sqrt{3} }{2} *a^2$

7 0

3 years ago

Read 2 more answers

Evaluate x3 - x + x0 for x = -1.

8090 [49]

Substitute (-1)3-(-1)+(-1)0
multiply (-3)+1+0
simplify. -2

Final Answer: -2

3 0

2 years ago

You have to solve the problems starting at the "begin" and make your way up to the "end". The correct answer is one of the probl

Thepotemich [5.8K]

Begin
3x-4=14
3x=18
x=6

.5n=1.25
n=2.5

x+1=2x-2
x+3=2x
3=x

x+3=10
x=7

6x-13=-11
6x=2
x=1/3

p=4-p
2p=4
p=2

End

6 0

2 years ago

Read 2 more answers

The value 5 is an upper bound for the zeros of the function shown below.

Mice21 [21]

Answer:

The given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.

Step-by-step explanation:

Given

$f\left(x\right)\:=\:x^2\:+\:x^3\:-\:11x^2\:-\:9x\:+\:18$

We know the rational zeros theorem such as:

if $x=c$ is a zero of the function $f(x)$ ,

then $f(c) = 0$ .

As the $f\left(x\right)\:=\:x^2\:+\:x^3\:-\:11x^2\:-\:9x\:+\:18$ is a polynomial of degree $4$ , hence it can not have more than $4$ real zeros.

Let us put certain values in the function,

$f(5) = 448$ , $f(4) = 126$ , $f(3) = 0$ , $f(2) = -20$ ,

$f(1) = 0$ , $f(0) = 18$ , $f(-1) = 16$ , $f(-2) = 0$ , $f(-3) = 0$

From the above calculation results, we determined that $4$ zeros as

$x = -3, -2, 1,$ and $3$ .

Hence, we can check that

$f(x) = (x+3)(x+2)(x-1)(x-3)$

Observe that,

$for x > 3$ , $f(x)$ increases rapidly, so there will be no zeros for $x>3$ .

Therefore, the given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.

5 0

3 years ago