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3 years ago

A cheetah runs at a speed of 180 kmph. It covers a distance of 400 m in:

Mathematics

1 answer:

Nutka1998 [239]3 years ago

4 0

Answer:

8 second

Step-by-step explanation:

400/180,000 × 3,600 second =

1,440,000/180, 000 = 8 second

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Evaluate the expression (3-2i)(3-2i) and write the result in the form a+bi ,where a is the real part of your answer and bis the

Maslowich

Answer:

a = 5, b = -12

Step-by-step explanation:

By basic factoring, you get 9 - 12i + $4i^2$ . Since $i^2$ is simply -1, the expression evaluates to 9 - 12i - 4, which is 5 - 12i.

4 0

2 years ago

What is the measurement of x in the equation. 8x+5x+2x-11+6=180

34kurt

Answer:

x = 37/3 = 12.333

Step-by-step explanation:

Step 1 :

Pulling out like terms :

1.1 Pull out like factors :

15x - 185 = 5 • (3x - 37)

Equation at the end of step 1 :

Step 2 :

Equations which are never true :

2.1 Solve : 5 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation :

2.2 Solve : 3x-37 = 0

Add 37 to both sides of the equation :

3x = 37

Divide both sides of the equation by 3:

x = 37/3 = 12.333

4 0

3 years ago

The area of the triangle formed by x− and y− intercepts of the parabola y=0.5(x−3)(x+k) is equal to 1.5 square units. Find all p

Juliette [100K]

Check the picture below.

based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).

and that will give us two x-intercepts, at x = 3 and x = k.

since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.

now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.

let's find the y-intercept by setting x = 0 now,

$\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)$

$\bf \cfrac{3}{2}=\cfrac{3+k}{2}\left( -\cfrac{3k}{2} \right)\implies \stackrel{\textit{multiplying by }\stackrel{LCD}{2}}{3=\cfrac{(3+k)(-3k)}{2}}\implies 6=-9k-3k^2 \\\\\\ 6=-3(3k+k^2)\implies \cfrac{6}{-3}=3k+k^2\implies -2=3k+k^2 \\\\\\ 0=k^2+3k+2\implies 0=(k+2)(k+1)\implies k= \begin{cases} -2\\ -1 \end{cases}$

now, we can plug those values on A = (1/2)bh,

$\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5$