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3 years ago

I really need help if you explain I will give brainliest

Mathematics

1 answer:

Troyanec [42]3 years ago

6 0

Since the ratio of sides is 4 to 3, the proportion is:

$\frac{4}{3} = \frac{12}{x}$

x = 9

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Work out the value of 2a + ay when a= 5 and y= -3

inn [45]

Just plug in a = 5, y = -3 in that!

2a + ay = 2(5) + 5(-3) = 10 -15 = -5
thats it!

5 0

3 years ago

Read 2 more answers

Find the following algebra, I am in class and plz give explanation

Lelechka [254]

1) Solving $4a+4b-ac-bc$ we get $(a+b)(4-c)$

2) Solving $16a^4-8b$ we get $8(2a^4-b)$

3) Solving $x^2-8x+15$ we get $(x-3)(x-5)$

Step-by-step explanation:

We need to solve the following expressions:

1) $4a+4b-ac-bc$

We will solve this expression by factoring common terms:

$4a+4b-ac-bc$

Factoring 4 from 4a+4b and -c from -ac-bc

$4(a+b)-c(a+b)$

Taking common term

$(a+b)(4-c)$

2) $16a^4-8b$

We will solve this expression by factoring common terms:

$16a^4-8b$

Taking 8 as common term:

$8(2a^4-b)$

3) $x^2-8x+15$

We can solve the expression by breaking the middle term into 2

$x^2-3x-5x+15\\Finding\,\,common:\\=x(x-3)-5(x-3)\\=(x-3)(x-5)$

Keywords: Finding factors

Learn more about Finding factors at:

#learnwithBrainly

6 0

3 years ago

A ball is thrown vertically in the air with a velocity of 95 ft/s. The ball is at a height of 120 ft.

sattari [20]

Answer:

The ball is at a height of 120 feet after 1.8 and 4.1 seconds.

Step-by-step explanation:

The statement is incomplete. The complete statement is perhaps the following "A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round to the nearest tenth of a second."

Since the ball is launched upwards, gravity decelerates it up to rest and moves downwards. The position of the ball can be determined as a function of time by using this expression:

$y = y_{o} + v_{o}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$y_{o}$ - Initial height of the ball, measured in feet.

$v_{o}$ - Initial speed of the ball, measured in feet per second.

$g$ - Gravitational constant, equal to $-32.174\,\frac{ft}{s^{2}}$ .

$t$ - Time, measured in seconds.

Given that $y_{o} = 0\,ft$ , $v_{o} = 95\,\frac{ft}{s}$ , $g = -32.174\,\frac{ft}{s^{2}}$ and $y = 120\,ft$ , the following second-order polynomial is found:

$120\,ft = 0\,ft + \left(95\,\frac{ft}{s} \right)\cdot t +\frac{1}{2}\cdot \left(-32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}$

$-16.087\cdot t^{2} + 95\cdot t -120 =0$

The roots of this polynomial are, respectively:

$t_{1} \approx 4.075\,s$ and $t_{2} \approx 1.831\,s$ .

Both roots solutions are physically reasonable, since $t_{1}$ represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas $t_{2}$ represents the instant when the ball the same height after reaching maximum height.