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2 years ago

Please I need help of this one :)

Mathematics

1 answer:

Harlamova29_29 [7]2 years ago

5 0

Answer:

128.605 m²

Step-by-step explanation:

Find the areas of the top triangle, and the bottom triangle and add them together.

The top triangle consists of 2 smaller triangles. Look at the larger one, the height is 4.93 and the base is 17. The area is...

A = (1/2)(17)(4.93)

A = 41.905 m²

The bottom triangle consists of 2 smaller triangles. Look at the larger one, the height is 10.2 and the base is 17. The area is...

A = (1/2)(17)(10.2)

A = 86.7 m²

The total area is 41.905 m² + 86.7 m² = 128.605 m²

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Consider the equation 2(ax+3)=4x−3(4x+8). For what value of a is there no solution to the equation?

erastovalidia [21]

Answer:

the value of $a$ such that there is no solution to the equation is $a = -4$ .

Step-by-step explanation:

Let first simplify the expression presented on statement. The equation has no solution if and only if $a$ is eliminated in the process and an absurd is the result (i.e. $0 = 7$ ).

$2\cdot (a\cdot x + 3) = 4\cdot x -3\cdot (4\cdot x + 8)$

$2\cdot a\cdot x +6 = 4\cdot x -12\cdot x -24$

$2\cdot a \cdot x+6=-8\cdot x-24$

$2\cdot a\cdot x +8\cdot x = -30$

$2\cdot (a+4)\cdot x = -30$

$(a+4)\cdot x = -15$

To obtain an absurd, we need that $a+4 = 0$ . Hence, the value of $a$ such that there is no solution to the equation is:

$a = -4$

Let prove the certainty of the result. We find that an absurd exist: ( $a = -4$ )

$0 = -15$

5 0

2 years ago

A soccer team scores 12 points in 5 matches. They scored the same number of points in each of their first 3 matches and one more

NeX [460]

Answer:

2 points the first 3 matches then 3 points the last 2 matches

Step-by-step explanation:

12 points in 5 matches and the first 3 are all the same then the other 2 increased by 1 point

They got 2 points the first 3 matches then 3 points the last 2 matches

2 + 2 + 2 (first 3 matches) = 6

3 + 3 (last 2 matches) = 6

6 + 6 = 12

7 0

2 years ago

Security A rounds every 2 minutes and 20 seconds in the gate 1. Security B rounds every 1 minute and 40 seconds in the gate 2. I

Elden [556K]

Answer:

$Time = 11\frac{2}{3}\ min$

Step-by-step explanation:

Given

$Security\ A = 2\ mins, 20\ secs$

$Security\ B = 1\ mins, 40\ secs$

Required

After how many minutes, will they round together

First, convert the given time to minutes

$Security\ A = 2\ mins, 20\ secs$

$Security\ A = 2\ mins+ 20\ secs$

$Security\ A = 2\ mins+ \frac{20}{60}\ min$

$Security\ A = 2\ mins+ \frac{1}{3}\ min$

$Security\ A = 2\frac{1}{3}\ min$

$Security\ B = 1\ mins, 40\ secs$

$Security\ B = 1\ mins+ 40\ secs$

$Security\ B = 1\ mins+ \frac{40}{60}\ min$

$Security\ B = 1\frac{2}{3}\ min$

So, we have:

$Security\ A = 2\frac{1}{3}\ min$

$Security\ B = 1\frac{2}{3}\ min$

List out the multiples of the time of both security personnel take round.

$Security\ A = 2\frac{1}{3}min,\ 4\frac{2}{3}min,\ 7min,\ 9\frac{1}{3}min,\ 11\frac{2}{3}min...$

$Security\ B = 1\frac{2}{3}min,\ 3\frac{1}{3}min,\ 5min,\ 6\frac{2}{3}min,\ 8\frac{1}{3}min,\ 10min\ ,11\frac{2}{3}min,...$

In the above lists, the common time is:

$Time = 11\frac{2}{3}\ min$

This implies that they go on round after $11\frac{2}{3}\ min$

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2 years ago

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3 years ago

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Nataly [62]

Answer:

$10,870

Step-by-step explanation:

we can multiply the initial value (27500) by (1-.06), or 0.94, raised to the 15th power

27500(.94)^15 = 10,870.52

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2 years ago