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2 years ago

Please helpp!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

kobusy [5.1K]2 years ago

6 0

Answer:

12 mpm

Step-by-step explanation:

Because he goes 3 miles in 36 mins so 36/3=12

Darina [25.2K]2 years ago

5 0

Answer:

A 1/12 mph

B 30 minutes

Step-by-step explanation:

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You are starting your own business and decided to take out a simple interest loan for 10,000 from your local bank. The interest

MAVERICK [17]

I think it’s 51.6% because 8.3% interest at the rate of 6 years is just 8.3 multiplied by 6 which is 51.6

7 0

1 year ago

Which value of x is in the domain of f(t) = squared x - 7

zloy xaker [14]

Given:

The given function is:

$f(x)=\sqrt{x-7}$

To find:

The value of x that is in the domain.

Solution:

Domain is the set of input values.

We have,

$f(x)=\sqrt{x-7}$

We know that the square root is defined for non negative values. So,

$x-7\geq 0$

$x\geq 0+7$

$x\geq 7$

Thus, the domain of the given function is all real number that are greater than or equal to 7.

In the given options 0, -3, 6 are less than 7 but 8 in option A is the only value that is greater than 7. So, $x=8$ is in the domain of the given function.

Therefore, the correct option is A.

8 0

2 years ago

A chord is 10cm from the centre of a circle of radius 15cm. Find the length of the chord.

almond37 [142]

Answer: 21.44 cm

Step-by-step explanation:

Given

Radius r=15 cm

distance of chord from the center is 10 cm

From the figure, the length of the chord is AC

Applying Pythagoras

$\Rightarrow AO^2=OB^2+AB^2\\\Rightarrow AB^2=15^2-10^2\\\Rightarrow AB^2=225-100=115\\\Rightarrow AB=\sqrt{115}=10.72\ cm$

The length of the chord is $AC=2AB=21.44\ cm$

5 0

2 years ago

The following proof shows an equivalent system of equations created from another system of equations. Fill in the missing reason

Nookie1986 [14]

Answer:

<h2>A.Multiplication Property of Equality</h2>

Step-by-step explanation:

$\left\{\begin{array}{ccc}2x+2y=14\\-x+y=5&\text{\underline{multiply both sides by 2}}\end{array}\right\\\\\boxed{\underline{+\left\{\begin{array}{ccc}2x+2y=14\\-2x+2y=10\end{array}\right}}\qquad\text{add both sides of the equations}\\.\qquad\qquad4y=24\qquad\text{divide both sides by 4}\\.\qquad\qquad y=6\\\\\text{put the value of y to the second equation:}\\\\-x+6=5\qquad\text{subtract 6 from both sides}\\-x=-1\qquad\text{change the signs}\\x=1$