FInd the unknown value of: 18 = 3 + x x = ?

PLEASE ANSWER ASAP! I'M GIVING A LOT OF POINTS!

nikdorinn [45]

In this question you need to isolate j. First you multiply 11 on both sides to make 3 + j not a fraction anymore.

3 + j = 22.
Then you isolate j by subtracting 3 on both sides since you must cancel the 3's out to isolate j. Then you get an answer of j = 19

7 0

2 years ago

Help pleaseeee Hajsbdheuebebbdhdjwiwkwkwjdfbfjdjdje Idk what to do

GREYUIT [131]

Answer:

Step-by-step explanation: square root both numbers so 55 square root is 7.41 and he square root of 16 is 4 now divide them so 7.41 divided by 4 = to 1.85 there fore 1.85 is ur answer

7 0

2 years ago

Help! Solve this for the points

stiv31 [10]

Formula for slope is M=y2-y1/x2/x1
Take two points from the graph and insert them into the formula. 13,32 and 7,33 M=32-33/13-7 which equals -1/6 this is already simplified I believe

Answer: -1/6

Sorry if anything is incorrect though.

4 0

6 months ago

Sarah is going to New York for spring break with her friends they have $130 to spend on going to the baseball game the tickets c

anyanavicka [17]

They can buy 4 tickets, with a little bit left over.

3 0

3 years ago

Read 2 more answers

As a bowl of soup cools, the temperature of the soup is given by the twice-differentiable function H for 0<img src="https://tex.

Aleks04 [339]

The rate of change of temperature with time at a point in time is given by

the derivative of the function for the temperature of the soup.

The correct responses are;

a) H'(5) is approximately -2.6 degrees Celsius per minute.

b) Yes

c) The equation for the line tangent is y = -3.6·t + 90.8
The approximate value of C(5) is 72.8 °C

d) The rate of change of the temperature of the soup at t = 3 minutes is -3.6 degrees Celsius per minute.

Reasons:

a) From the data in the table, we have;

The approximate value of H'(5) is given by the average value of the rate of

change of the temperature with time between points, t = 3, and t = 8

Therefore;

$\displaystyle H'(5) = \mathbf{\frac{H(8) - H(3)}{8 - 3}}$

Which gives;

$\displaystyle H'(5) = \frac{80 - 67}{8 - 3} = \mathbf{2.6}$

Therefore, H'(5) = -2.6°C per minute

b) Given that the function is twice differentiable over the interval, 0 ≤ t ≤ 12, the function for the change in temperature is continuous in the interval 0 ≤ t ≤ 12

At t = 0, H(0) = 90 °C

At t = 12, H(12) = 58 °C

58 °C < 60 < 90 °C

Therefore, there exist a temperature, of 60 °C between 90° C and 58 °C

c) The given derivative of C is, $C'(t) = \mathbf{-3.6 \cdot e^{-0.05 \cdot t}}$

At t = 3, we have;

$The \ slope \ at \ t = 3 \ is \ C'(3) = -3.6 \cdot e^{-0.05 \times 3} \approx -3.1$

Therefore, we have;

y - 80 ≈ -3.1 × (x - 3)

The equation for the tangent is; y = -3.6 × (x - 3) + 80

y = -3.6·x + 10.8 + 80 = -3.6·x + 90.8

The equation for the tangent is; y = -3.6·x + 90.8

The value of C(5) is approximately, C(5) ≈ -3.6 × 5 + 90.8 = 72.8

C(5) ≈ 72.8°

d) Based on the the model above, the rate at which the temperature of the

soup is changing at t = 3 minutes is -3.6 degrees per minute.

Learn more about calculus and concepts here:

brainly.com/question/20336420

8 0

2 years ago