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3 years ago

What number is 18% of 250

Mathematics

2 answers:

vladimir2022 [97]3 years ago

7 0

18 over 100 = x over 250, x= 45. 45 is ur answer.

Alchen [17]3 years ago

5 0

18% of 250 is 45%

Change 18 into a decimal which is: 0.18.
0.18 × 250 = 45.

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Which system of equations can you use to find the roots of the equation?<br> x3 – 10x = x2 – 6

vivado [14]

$\bf \stackrel{y}{x^3-10x}=\stackrel{y}{x^2-6}\implies \begin{cases} y=x^3-10x\\ y=x^2-6 \end{cases}$

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2 years ago

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Refer to the triangle below. If a = 4 and b = 6, solve for c. show work Also, give the unsimplified radical form answer and then

9966 [12]

Look at the image below where I labeled the sides

To solve this you must use Pythagorean theorem:

$a^{2} +b^{2} =c^{2}$

a and b are the legs (the sides that form a perpendicular/right angle)

c is the hypotenuse (the side opposite the right angle)

In this case...

a = 4

b = 6

c = unknown

^^^Plug these numbers into the theorem

$4^{2} +6^{2} =c^{2}$

simplify

16 + 36 = $x^{2}$

52 = $x^{2}$

To remove the square from x take the square root of both sides to get you...

√52 = x

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7.21

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3 years ago

Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

——————————

Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

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2 years ago

Which ratios are equivalent to 3:1?

trapecia [35]

Answer:

9:3;27:9;87:27

Step-by-step explanation:

The list goes on but hope that answers your question

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2 years ago

A line has a slope of –5 and a y-intercept of (0, 3). What is the equation of the line that is perpendicular to the first line a

neonofarm [45]

Slope-intercept form:
y=mx+b
m=slope
b=y-intercept

Data of the first line:
m=-5
b=y-intercept=3 (y-intercept=it is the value of "y" when x=0)

y=-5x+3

A line perpendicular to the line y=mx+b will have the following slope:
m`=-1/m

Therefore: the line perpendicular to the line y=-5x+3 will have the following slope:
m´=-1/(-5)=1/5

Point-slope form of a line: we need a point (x₀,y₀) and the slope (m):
y-y₀=m(x-x₀)

We know, the slope (m=1/5) and we have a point (3,2) therefore:
y-y₀=m(x-x₀)
y-2=1/5(x-3) (point-slope form)
y-2=(1/5)x-3/5
y=(1/5)x-3/5+2
y=(1/5)x+7/5 (slope-intercept form)

Answer: the line perpendicular to the first line will be: y=(1/5)x+7/5

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2 years ago

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