Please answer correctly !!!!!!!!!! Will Mark brainliest !!!!!!!!!!!!

Mathematics

1 answer:

nydimaria [60]3 years ago

3 0

Answer:

Step-by-step explanation:

The discriminant is what's under the square root sign in the quadratic equation. The equation for the discriminant is $b^{2}-4ac$ , where b is the coefficient of x, a is the coefficient of $x^{2}$ , and c is the number with no variable attatched to it. If we plug in the numbers ( $17^{2} -4*4*3$ ) it gives you 241, which is the discriminant. Since 241 is more than zero, it has 2 zeros. If the discriminant was 0, there'd be 1 zero, and less then zero there would be zero zeros.

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In Scott’s backyard, he has a windmill that makes 5 ¾ of a turn every hour. How many turns does it make for 1 ½ hours?

DanielleElmas [232]

Answer: 8 5/8 turns

Step-by-step explanation:

Hi, to answer this question we have to simply to multiply the number of turns that the windmill makes per hour (5 3/4) by the time asked (1 ½ hours).

Mathematically speaking:

5 3/4 x 1 1/2 = (5x4+3)/4 x (1x2+1)/2 = 23/4 x 3/2= 69/8 or 8 5/8 (mixed form)

In conclusion, the windmill makes 8 5/8 turns in 1 1/2 hours.

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3 years ago

An electrician can wire on average 4.5 houses in a week. How many months will it take her to wire 55 houses if she wires the sam

Brrunno [24]

Just do 4.5 x4.5 and get 20.25 then divide that by 55 and get the answer of 2.7 months. Hope this helps

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3 years ago

What number should be added to both sides of the equation to complete the square? x2 8x = 4

Levart [38]

You would take half of the x term...and square it
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6 0

3 years ago

Read 2 more answers

What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)

Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = $L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\$

Now to solve the integral on the right hand side we shall use Integration by parts Taking $7t^{3}$ as first function thus we have

$\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\$

Again repeating the same procedure we get

$=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\$

Again repeating the same procedure we get

$\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\$