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horsena [70]
2 years ago
5

2⁴+(2.75+1.75)÷0.9 fast pls helppppp

Mathematics
1 answer:
alisha [4.7K]2 years ago
5 0

Answer:

21

Step-by-step explanation:

First, add the parenthesis

2^4 + 4.50 ÷ 0.9

Second, divide 4.50 by 0.9

2^4 + 5

Third, get rid of the ^4

(2x2x2x2)

16 + 5

21

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What is the equation of the line in y=mx+b format? BRAINLIST WILL BE REWARDED
harkovskaia [24]

Answer:

y = 2/5x - 1

Step-by-step explanation:

There are a couple of different ways to figure this out, but since you have a graph, the fastest way is to count the slope, or m in the formula, by counting the rise (units up) over run (units to the right), which is 2/5. Then, find the value of b by finding the y-intercept on the graph, or -1.

3 0
3 years ago
50pt! The function g(x)=x^2 is transformed to obtain function h: h(x)=g(x-3).
Vlad1618 [11]

Answer:

It's the Option A.

Step-by-step explanation:

h(x)  = g(x - 3)

The -3 moves the whole graph of g(x)  3 units horizontally to the right.

5 0
3 years ago
Tell whether the sequence is arithmetic if it is identify the common difference -9,-17,-26,-33
aleksklad [387]
No it is not, there is no common difference. <span />
8 0
3 years ago
Read 2 more answers
PLEASE I NEED HELP ​
NeTakaya

Answer:

D) y=-1/2x+4

Step-by-step explanation:

D) y=-1/2x+4

In order to be parallel, it has to have the same slope (-1/2).  Also, if you plug in (-2,5) for x and y, it does pass through those coordinates.

8 0
3 years ago
The position of an object along a vertical line is given by s(t) = −t3 + 3t2 + 7t + 4, where s is measured in feet and t is meas
saw5 [17]

Answer:

The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

Step-by-step explanation:

Given : The position of an object along a vertical line is given by s(t) = -t^3+3t^2+7t +4, where s is measured in feet and t is measured in seconds.

To find : What is the maximum velocity of the object in the time interval [0, 4]?

Solution :

The velocity is rate of change of distance w.r.t time.

Distance in terms of t is given by,

s(t) = -t^3+3t^2+7t +4

Derivate w.r.t. time,

v(t)=s'(t) = -3t^2+6t+7

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

x=-\frac{b}{2a}

Where, a=-3, b=6 and c=7

t=-\frac{6}{2(-3)}

t=-\frac{6}{-6}

t=1

As 1 lie between interval [0,4]

Substitute t=1 in the function,

v(t)= -3(1)^2+6(1)+7

v(t)= -3+6+7

v(1)=10

Th maximum velocity is 10 ft/s.

Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

8 0
3 years ago
Read 2 more answers
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