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Alenkasestr [34]
3 years ago
6

Geometry question please help

Mathematics
1 answer:
Bess [88]3 years ago
3 0
Your answer would be C 


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Can someone help me with these 4 questions please :) .
butalik [34]

1.

60² + 30² = 4500²

Opposite corner = 67.08 ft.

//

2.

6² + 8² = 100²

The wire must be 10 ft.

//

3.

7² + 14² = 245²

The rope is 15.65 ft.

//

4.

180² + 300² = 122400²

He ran 349.85 ft.

//

I'm not too sure about question 3.

8 0
3 years ago
In training, Jayden does reps where he bench presses 65 kilograms. How many reps does he have to do to bench press a total of 1
WITCHER [35]
He would have to do 15.384 reps
5 0
3 years ago
How much money will you have at the end of one year if interest is compounded semiannually at 10% on a $600 deposit?
Masja [62]
The formula required to answer this is:
A=P(1+\frac{r}{n})^{nt}
where  A is the amount of the principal P after n compounding periods per year at r interest rate (as a decimal fraction).
Plugging in the given values, we get:
A=600(1+\frac{0.1}{2})^{2\times1}=661.5
Rounding to the nearest dollar, we get $662.00. Therefore a is the best answer choice.

6 0
4 years ago
What number has a 9 with a value ten times as many as the 9 in 39,154
Tomtit [17]

Answer:

Any number with 9 in the ten-thousands place. 90,000 is one such number.

Step-by-step explanation:

The 9 in 39,154 is in the thousands place. Its value is 9,000. In order for the 9 in a number to have a value 10 times that, or 90,000, the 9 must be in the ten-thousands place.

There are an infinite number of such numbers. We suspect you have a list you are to choose from. Pick the number with 9 where it is in the number 90,000.

8 0
3 years ago
AABC has vertices at A(1, -9), B(8,0), and C(9,-8).
Rainbow [258]

Check the picture below.

~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{-9})\qquad B(\stackrel{x_2}{8}~,~\stackrel{y_2}{0}) ~\hfill AB=\sqrt{[ 8- 1]^2 + [ 0- (-9)]^2} \\\\\\ AB=\sqrt{7^2+(0+9)^2}\implies AB=\sqrt{7^2+9^2}\implies \boxed{AB=\sqrt{130}} \\\\[-0.35em] ~\dotfill

B(\stackrel{x_1}{8}~,~\stackrel{y_1}{0})\qquad C(\stackrel{x_2}{9}~,~\stackrel{y_2}{-8}) ~\hfill BC=\sqrt{[ 9- 8]^2 + [ -8- 0]^2} \\\\\\ BC=\sqrt{1^2+(-8)^2}\implies \boxed{BC=\sqrt{65}}

now, we could check for the CA distance, however, we already know that AB ≠ BC, so there's no need.

6 0
2 years ago
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