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3 years ago

6

A scale drawing of a rectangular basketball court measures 7 inches long and 4 inches wide. If the scale used for the drawing is

2.5 inches equals 5 yards, what is the actual length of the park?

1 answer:

gavmur [86]3 years ago

4 0

Answer:

14 yards long and 8 yards long

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Rewrite the equation in standard form 2x^2=5x+6

FrozenT [24]

Answer:

$2x^{2} =5x+6\\$

Step-by-step explanation:

8 0

2 years ago

I will give brainliest!

svetlana [45]

A copy of a line segment will have the same measure or length as the segment.

The true statement is (4) They are congruent.

From the question, we understand that a copy of the line segment is created.

This means that the new segment has the exact property as the original line segment.

Hence, both lines are congruent, and the true statement is (d)

Read more about line segments at:

brainly.com/question/3573606

6 0

2 years ago

3х – 6+1 = - 2x – 5+5х

marysya [2.9K]

Answer:0=12

Step-by-step explanation:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation:

3x-7=3x-5

Combine like terms:

3x-3x=7+5

0=12

4 0

3 years ago

Read 2 more answers

Rewrite each equation in vertex form by completing the square. Then identify the vertex.

ioda

ANSWER

Vertex form;

$y = 3( {x + \frac{3}{2} })^{2} - \frac{35}{4}$

Vertex

$V( - \frac{3}{2} , - \frac{35}{4} )$

EXPLANATION

Given:

$f(x) = 3 {x}^{2} + 9x - 2$

We complete the square as follows:

$y = 3( {x}^{2} + 3x) - 2$

$y = 3( {x}^{2} + 3x + \frac{9}{4} ) - 2 - 3 \times \frac{9}{4}$

The vertex form is:

$y = 3( {x + \frac{3}{2} })^{2} - \frac{35}{4}$

The vertex is

$V( - \frac{3}{2} , - \frac{35}{4} )$

6 0

3 years ago

Read 2 more answers

Consider the differential equation y'' − y' − 30y = 0. Verify that the functions e−5x and e6x form a fundamental set of solution

Alex Ar [27]

Answer:

Step-by-step explanation:

We have to take the derivatives for both functions and replace in the differential equation. Hence

for y=e^{-5x}:

$y(x)=e^{-5x}\\y'(x)=-5e^{-5x}\\y''(x)=25e^{-5x}\\$

for y=e^{6x}:

$y(x)=e^{6x}\\y'(x)=6e^{6x}\\y''(x)=36e^{6x}\\$

Now we replace in the differential equation y'' − y' − 30y = 0

for y=e^{-5x}:

$25e^{-5x}+5e^{-5x}-30e^{-5x}=0\\25+5-30=0$

for y=e^{6x}:

$36e^{6x}-6e^{6x}-30=0\\36-6+30=0$

Now, to know if both function are linearly independent we calculate the Wronskian

$W(f,g)=fg'-f'g$

$W(e^{-5x},e^{6x})=(e^{-5x})(6e^{6x})-(-5e^{-5x})(e^{6x})\neq 0$

I hope this is useful for you

Best regard

7 0

3 years ago