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goldfiish [28.3K]
2 years ago
13

Which expression is equivalent to (5x - 4)2?

Mathematics
2 answers:
pogonyaev2 years ago
8 0

Answer:

10x-8

Step-by-step explanation:

Apply the distributive law: a(b-c)=ab-ac

a=2,b=5x.c=4

2x5x-2x4

simplify

=10x-8

Bond [772]2 years ago
7 0

Answer:

Step-by-step explanation:

the answer is 25x^2-40x+16

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Jorge bikes to school each day. If he can travel 33 miles in 11 hours, how fast does he travel in one
Eddi Din [679]

33 miles divided by 11 hours will give you the mile per hour traveled.

Answer: 3 miles per hr

3 0
2 years ago
Someone help me out here.
Vladimir [108]
Cremini:
$11.25 for 2/3 lb
2/3 lb = $11.25
1 lb = 11.25 ÷ 2/3 = $16.88 (nearest cent)

Chanterelle:
$7.99 for 1/2 lb
1/2 lb = $7.99
1 lb = $7.99 x 2 = $15.98

Answer:  Chanterelle costs less. It costs $0.90 less than Cremini.

4 0
3 years ago
Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
grandymaker [24]
(2x+1)^{\cot x}=\exp\left(\ln(2x+1)^{\cot x}\right)=\exp\left(\cot x\ln(2x+1)\right)=\exp\left(\dfrac{\ln(2x+1)}{\tan x}\right)

where \exp(x)\equiv e^x.

By continuity of e^x, you have

\displaystyle\lim_{x\to0^+}\exp\left(\dfrac{\ln(2x+1)}{\tan x}\right)=\exp\left(\lim_{x\to0^+}\dfrac{\ln(2x+1)}{\tan x}\right)

As x\to0^+ in the numerator, you approach \ln1=0; in the denominator, you approach \tan0=0. So you have an indeterminate form \dfrac00. Provided the limit indeed exists, L'Hopital's rule can be used.

\displaystyle\exp\left(\lim_{x\to0^+}\dfrac{\ln(2x+1)}{\tan x}\right)=\exp\left(\lim_{x\to0^+}\dfrac{\frac2{2x+1}}{\sec^2x}\right)

Now the numerator approaches \dfrac21=2, while the denominator approaches \sec^20=1, suggesting the limit above is 2. This means

\displaystyle\lim_{x\to0^+}(2x+1)^{\cot x}=\exp(2)=e^2
7 0
3 years ago
I need help now plz !
My name is Ann [436]

Answer:D

Step-by-step explanation:

students playing trombone=12+10+7+16

Students playing trombone=45

8 0
3 years ago
Find the point on the line y = 2x + 3 that is closest to the origin.
ch4aika [34]
1)We have to make a line perpendicular to y=2x+3, passing through the point (0,0).

y=2x+3   
m=solpe=2

The slope of the line perpedicular to y=2x+3 is m´
m´=-1/m
m´=-1/2

Point slope form:
y-y₀=m(x-x₀)
(0,0)
m=-1/2

y-0=-1/2(x-0)
y=-x/2

Therefore, the line perpendicular to y=2x+3 is y=-x/2.

2)The point on the line y=2x+3 tha is closest to the origin is the point of intersection of the two lines.

y=2x+3
y=-x/2

We can solve this system of equations by equalization method.
2x+3=-x/2
least common multiple=2
4x+6=-x
4x+x=-6
5x=-6
x=-6/5  (=-1.2)

y=-x/2
y=-(-6/5)/2=6/10=3/5    (=0,6)

Therefore: the point of the line y=2x+3 that is closest to the origin is:
 (-1.2 , 0.6)

3 0
3 years ago
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