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2 years ago

8

Please help me I need to turn in tomorrow please help!!!!

2 answers:

Sphinxa [80]2 years ago

4 0

Answer:

5/7 which is option c the correct answer.

butalik [34]2 years ago

4 0

Answer:

C. 5/7 is the correct answer

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Write an equation for the statement and then solve

Lunna [17]

Answer:

16. 30 * (30+12)/4*(L)*20

17. Volume = 6300 * Length of rectangular prism

Step-by-step explanation:

The width of a rectangular prism is <u>30 cm</u>. This is <u>12 more than one-fourth of the length</u>. Find the volume of the prism, given the <u>height is 20 cm</u>.

Let L = length of rectangular prisim W = width and H = height

16.

Volume of a rectangular prism is width * length * height

30 * (30+12)/4*(L)*20

17.

= 30 * (30+12)1/4(L)*20

= 30 * (42/4)*L * 20

= 600 * 10.5 * L

= 6300 * L

Volume = 6300 * Length of rectangular prism

6 0

2 years ago

Solve step by step solution then only i can do it plxx

Anuta_ua [19.1K]

Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°

So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle

So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

3 0

2 years ago

17 out 22 what percentage is correct

SVETLANKA909090 [29]

The answer is <span>77.2727272727273% hopw this help

</span>

5 0

2 years ago

arah mailed her computer to Store B for repairs. Her bill was: Total Cost of Computer Repair = $1,444.50 Shipping = 2 percent of

nataly862011 [7]

1415.61 IS HER TOTAL BILL BEFORE SHIPPING. 1415.61X .15=212.34

1415.61 REPAIR
28.89 SHIPPING
212.34 TIP
1656.84 TOTAL

7 0

3 years ago

Read 2 more answers

Give the following equations determine if the lines are parallel perpendicular or neither

GaryK [48]

In order to determine whether the equations are parallel, perpendicular, or neither, let's simply each equation into a slope-intercept form or basically, solve for y.

Let's start with the first equation.

$\frac{6x-5y}{2}=x+1$

Cross multiply both sides of the equation.

$6x-5y=2(x+1)$ $6x-5y=2x+2$

Subtract 6x on both sides of the equation.

$6x-5y-6x=2x+2-6x$ $-5y=-4x+2$

Divide both sides of the equation by -5.

$-\frac{5y}{-5}=\frac{-4x}{-5}+\frac{2}{-5}$ $y=\frac{4}{5}x-\frac{2}{5}$

Therefore, the slope of the first equation is 4/5.

Let's now simplify the second equation.

$-4y-x=4x+5$

Add x on both sides of the equation.

$-4y-x+x=4x+5+x$ $-4y=5x+5$

Divide both sides of the equation by -4.

$\frac{-4y}{-4}=\frac{5x}{-4}+\frac{5}{-4}$ $y=-\frac{5}{4}x-\frac{5}{4}$

Therefore, the slope of the second equation is -5/4.

Since the slope of each equation is the negative reciprocal of each other, then the graph of the two equations is perpendicular to each other.

5 0

10 months ago