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3 years ago

What is 95-65= and then divide it by five then multiply it by ten

Mathematics

2 answers:

Marina CMI [18]3 years ago

4 0

(95-65)/5*10=60
because 95-65=30
30/5=6
6*10=60

Natalka [10]3 years ago

4 0

95 - 65 = 30

30/5 = 6

6 × 10 = 60

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vitfil [10]

Answer:

Step-by-step explanation:

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b = 12(leg)

z = ?

z= $\sqrt{a^2-b^2}$ = 16

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3 years ago

The diameter of a circle has endpoints A(1, 3) and B(4, 7), what is the

lana [24]

Answer:

The circumference of the circle is 5π

Step-by-step explanation:

First we need to calculate the diameter

for this we have to know the distance between A(1, 3) and B(4, 7)

X = Xb - Xa

X = 4 - 1

X = 3

Y = Yb - Ya

Y = 7 - 3

Y = 4

b - a = (3 , 4)

now that we have the distance of x and r we can calculate its distance using Pythagoras

h = hypotenuse

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we replace the known values

h² = (3)² + (4)²

h² = 9 + 16

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The diameter of the circle is 5

Now we need to calculate the circumferencefor this we need to use the circumferenc formula of a circle:

c = circumference

d = diameter = 5

π = pi

c = π * d

we replace with the known values

c = 5 * π

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The circumference of the circle is 5π

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3 years ago

9,6,3,0,-3,-6 work out the nth term (it decrease by the same amount each time)

Shkiper50 [21]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

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kvv77 [185]

A) 6 4/7

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4 years ago

Find the equation, (f(x) = a(x - h)2 + k), for a parabola containing point (2, -1) and having (4, -3) as a vertex. What is the s

Nataliya [291]

Answer:

$f(x)=\frac{1}{2}x^2-4x+5$

Step-by-step explanation:

A parabola is written in the form

$f(x)=a((x-h)^2+k)$ (1)

where:

$h$ is the x-coordinate of the vertex of the parabola

$ak$ is the y-coordinate of the vertex of the parabola

$a$ is a scale factor

For the parabola in the problem, we know that the vertex has coordinates (4,-3), so we have:

$h=4$ (2)

$ak=-3$

From this last equation, we get that $a=\frac{-3}{k}$ (3)

Substituting (2) and (3) into (1) we get the new expression:

$f(x)=-\frac{3}{k}((x-4)^2+k) = -\frac{3}{k}(x-4)^2 -3$ (4)

We also know that the parabola contains the point (2,-1), so we can substitute

x = 2

f(x) = -1

Into eq.(4) and find the value of k:

$-1=-\frac{3}{k}(2-4)^2-3\\-1=-\frac{3}{k}\cdot 4 -3\\2=-\frac{12}{k}\\k=-\frac{12}{2}=-6$

So we also get:

$a=-\frac{3}{k}=-\frac{3}{-6}=\frac{1}{2}$

So the equation of the parabola is:

$f(x)=\frac{1}{2}((x-4)^2 -6)$ (5)

Now we want to rewrite it in the standard form, i.e. in the form

$f(x)=ax^2+bx+c$

To do that, we simply rewrite (5) expliciting the various terms, we find:

$f(x)=\frac{1}{2}((x^2-8x+16)-6)=\frac{1}{2}(x^2-8x+10)=\frac{1}{2}x^2-4x+5$

6 0

3 years ago