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3 years ago

11

Tim wants to buy a car. The first car is $8495 and has 20%off. The second car is $8142 and has 15%off. Which car is cheapest and

by how much??
Did anybody know this? Help me

1 answer:

finlep [7]3 years ago

6 0

The first car is gonna by the cheapest by 124.70

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Help plz thx I'm new here

iris [78.8K]

3. 20-(5x2). The answer is 10. [20-(5x2) -> 20-10=10]
4. B (Parenthesis are needed) and C (Start with simplifying the bottom).
5. 6x(9-3). The answer is 36. [6x(9-3) -> 6x6=36]
6. 10 divided by the difference of 6 and 4.
7. B
8. (5+7)x3 -> 12x3=36

4 0

3 years ago

Keiko drove 225 miles using 10 gallons of gas. At this rate, how many gallons of gas would she need to drive 441 miles?

lidiya [134]

Answer:

19.6 gallons

Step-by-step explanation:

divide 225 by 10 and you get 22.5 this means you are getting 22.5 miles per gallon so all you have to do is divide 441 by 22.5 and you get 19.6 so you are going to need 19.6 gallons to drive 441 miles

6 0

3 years ago

Solve the system of equations.

Brilliant_brown [7]

Answer: B

Step-by-step explanation:

Setting the equations equal to each other,

-5x-8 = 4x+1

-8 = 9x + 1

-9 = 9x

x = -1, meaning y=-5(-1)-8 = -3.

7 0

3 years ago

Juanita is 13 years older than her cousin. The sum of their ages is no less than 103 years. Enter an inequality that can be used

Bond [772]

Please see below solution:

y = x + 13<span>
y + x ≥ 103
x + 13 + x ≥ 103
Inequality is:
2 x + 13 ≥ 103
2 x ≥ 103 - 13
2 x ≥ 90
x ≥ 90 : 2
x ≥ 45
<span>The youngest age Juanita`s cousin can be is 45.</span></span>

6 0

4 years ago

Suppose that ff is a differentiable function on the real line, we have −2≤f′(x)≤4−2≤f′(x)≤4 for all real numbers xx and f(2)=4f(

MariettaO [177]

Answer:

(a)-6 and 24

(b)-12 and 12.

Step-by-step explanation:

Using the mean value theorem,

Suppose that f is a function which is continuous on a closed interval [a, b] and differentiable on an open interval (a, b), then there

exists a number c in (a, b) so that

$f^{'}(c)= \frac{f(b)-f(a)}{b-a}$

for every closed interval [a,b].

−2≤f′(x)≤4 and f(2)=4

First, we determine the largest and smallest possible value for f(7).

In the Interval [2,7],

$f^{'}(x)= \frac{f(7)-f(2)}{7-2}$

Since −2≤f′(x)≤4

$-2$\leq$\frac{f(7)-f(2)}{5}$\leq$4$

Recall that: f(2)=4

$-2$\leq$ \frac{f(7)-4}{5}$\leq$4$

$-2X5$\leq$ f(7)-4$\leq$4X5$

$-10+4$\leq$ f(7)-4+4$\leq$20+4$

$-6$\leq$ f(7)$\leq$24$

The greatest and least value are 24 and -6 respectively.

Similarly for f(-2)

In the Interval [-2,2],

$f^{'}(x)= \frac{f(2)-f(-2)}{2-(-2)}$

Since −2≤f′(x)≤4

$-2$\leq$ \frac{f(2)-f(-2)}{4}$\leq$4$

Recall that: f(2)=4

$-2$\leq$ \frac{4-f(-2)}{4}$\leq$4$

$-2X4$\leq$ -f(-2)+4$\leq$4X4$

$-8-4$\leq$ -f(-2)-4+4$\leq$16-4$

$-12$\leq$ -f(-2)$\leq$12$

Dividing all through by negative

$-12$\leq$ f(-2)$\leq$12$

The greatest and least value are 12 and -12 respectively.

3 0

3 years ago