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3 years ago

I really need help with these problems

Mathematics

1 answer:

White raven [17]3 years ago

4 0

10.
$x^2+14x+48=x^2+6x+8x+48=x(x+6)+8(x+6)=\boxed{(x+6)(x+8)}$

11.
$\dfrac{t^2-4t-32}{t-8}=\dfrac{t^2-8t+4t-32}{t-8}=\dfrac{t(t-8)+4(t-8)}{t-8}\\\\=\dfrac{(t-8)(t+4)}{t-8}=t+4\\\\Answer:\boxed{t+4;\ t\neq8}$

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A car, initially at rest, accelerates at 2 meters/s/s. assuming the car starts from rest, how far will it travel in 10 s?

MrRa [10]

We shall use the formula for distance traveled:
s = ut + (1/2)at^2
where
u = initial velocity
t = time
a = acceleration.

Because
u = 0 (car starts from rest),
t = 10 s (travel tme),
a = 2 m/s^2 (acceleration), m
the di mstance traveled is
s = (1/2)*(2 m/s^2)*(10 s)^2 = (1/2)*2*100 = 100

Answer: 100 m

8 0

3 years ago

The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c

Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 $\rm{in^3}$

Step-by-step explanation:

The linearization or linear approximation of a function $f(x)$ is given by:

$f(x_0+dx) \approx f(x_0) + df(x)|_{x_0}$ where $df$ is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh$

Taking $R_0=1.5$ inches and $h=3$ inches and evaluating the partial derivatives we obtain:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh$

substituting the values and taking $dx=0.1$ and $dh=0.3$ inches we have:

$V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}$

Therefore the change in volume is estimated to be 17.20 $\rm{in^3}$

4 0

2 years ago

What is -8? PLEASE HELP ME

Mrac [35]

-8 IS A NEGATIVE INTEGER

6 0

3 years ago

Read 2 more answers

the cost of 5 cans of dog food is $4.35. at this price how much do 11 cans of dog food cost explain your reasoning. ASAP

mestny [16]

Answer:

Cost of 11 cans of Dog food is $9.57

Step-by-step explanation:

Cost of 5 cans of dog food = $4.35

So to find out the cost of 11 cans of dog food at the same price, we need to calculate the price of 1 can of dog food.

Calculation for finding out cost of 1 can of dog food

The price $4.35 which is given is the price of 5 cans

So to find the cost of 1 can, we need to divide the given price by 5

5 Cans of dog food = $4.35

1 can of dog food = 4.35 ÷ 5

1 can of dog food = $0.87

Now to find out the price of 11 cans, we will multiply the price of 1 can with 11

Cost of 11 cans = 11 × $0.87

= $9.57

6 0

2 years ago

Find the solution to the following system by substitution.

ASHA 777 [7]

Answer:

(5,15)

Step-by-step explanation:

5 0

2 years ago

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