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RUDIKE [14]
3 years ago
9

Find the greiple ofR(x) =( x ^3+10x^2 +32x+32)÷(X^2 + 2x-15)​

Mathematics
1 answer:
shtirl [24]3 years ago
6 0

Answer:

x3(+10x2 +32x +32) / (2 +30)

Step-by-step explanation:

(x3 +10x2 + 32x +32) / (x2 + 2 * 15)

x3(+10x2 +32x +32) / (2 +30)

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Find the measure of the indicated angle to the nearest degree
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\cos \theta =\dfrac{18}{23}\\\\\implies \theta = \cos^{-1}\left(\dfrac{18}{23} \right)=38.5^{\circ} \approx 39^{\circ}

7 0
2 years ago
What is two thirds of four fifths?
BabaBlast [244]

Answer:

The word "of" in the question means multiply, so you need to multiple these two fractions.

2/3 x 4/5 = 8/15

8 0
3 years ago
Read 2 more answers
Please help!! This is super important!!
MAVERICK [17]

Answer:

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3 years ago
Find all the values of k so that the given points are 29 square units apart. (-5, 5) and (k,0).
Varvara68 [4.7K]
Distance formula between 2 points (x1,y1) and (x2,y2)

D=\sqrt{(x1-x2)^{2}+(y1-y2)^{2}}

D=29 and one point is (-5,5) and the other is (k,0)
sub

29=\sqrt{(-5-k)^{2}+(5-0)^{2}}
29=\sqrt{(-5-k)^{2}+(5)^{2}}
square both sides
841=(-5-k)^{2}+(5)^{2}
841=(-5-k)^{2}+25
minus 25 both sides
816=(-5-k)^{2}
square root both sides don't forget positive and negative roots
+/-4√51=-5-k
add 5 to both sides
5+/-4√51=-k
times -1 both sides
-5+/-4√51=k




the possible values for k are
k=-5+4√51 or k=-5-4√51

7 0
3 years ago
The top of a ladder slides down a vertical wall at a rate of 2 feet per second. At the moment when the bottom of the ladder is 4
lord [1]

Answer:

The length of the ladder is 7.21 feet

Step-by-step explanation:

Let

y = Height of the wall

x = Distance between the wall and the ladder (bottom)

L = Length of the ladder

So, the given parameters are:

\frac{dy}{dt} = -2ft/s

When x = 4; \frac{dx}{dt} = 3ft/s

The ladder, the wall and the ground forms a right-angled triangle where the hypotenuse is the length of the ladder; So:

L^2 = x^2 + y^2

Differentiate with respect to time

2L\frac{dL}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}

The length of the ladder does not change with time. So:

2L*0 = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}

0 = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}

Rewrite as:

2x\frac{dx}{dt} =- 2y\frac{dy}{dt}

Divide both sides by 2

x\frac{dx}{dt} =- y\frac{dy}{dt}

Recall that:

\frac{dy}{dt} = -2ft/s and  x = 4; \frac{dx}{dt} = 3ft/s

So:

4 * 3 = -y * -2

12 = 2y

6 = y

y = 6

Substitute y = 6 and x =4 in: L^2 = x^2 + y^2

L^2 =4^2 + 6^2

L^2 =52

L = \sqrt{52

L = 7.21ft

4 0
3 years ago
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