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2 years ago

Write the number described by 1 ten 16 ones.

2 answers:

8 0

1 ten and 16 ones is 26, because 1 ten is 10 then 16 ones is 16 so 16+10=26. Hope that helps. If it did please click the thanks button and if you have nay questions please ask.

Alexxx [7]2 years ago

7 0

1 ten 16 ones=1*10+16*1=10+16=26

Hope this helps!

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Bonita is testing prototype model rocket engines. To be considered a successful launch, the rocket must reach a height of at lea

lbvjy [14]

Answer:

(0, 33) : Yes

(4.8, 30.5) : Yes

Step-by-step explanation:

height of at least 24 ft : y >= 24

horizontal distance of no more than 10 ft. : 0 <= x <= 10

The vertical height must be at least three times as high as the horizontal distance. No rocket should go higher than 33 ft.:

y >= 3x and y <= 33

So Constraints:

24 <= y <= 33

y >= 3x

0 <= x <= 10

Now check to see if those coordinate points meet all the constraints

(0, 33) : Yes

(4.8, 30.5) : Yes

(9, 26) : No (because 26 is less than (3 * 9) = 27)

(4, 36) : No (because 36 > 33, must be 24 <= y <= 33)

(2, 22): No (because 22 <24, must be 24 <= y <= 33)

5 0

2 years ago

Ashley invests $9,720 in a one-month money market account paying 3.16% simple annual interest and $8,140 in a two-year CD yieldi

tester [92]

For the first investment. A = P(1 + rt); where p = 9,720, r = 0.0316 and t = 1/12
A = 9720(1 + 0.0316/12) = 9720(1.0026) = $9,746

For the second investment,
A = 8140(1 + 0.0323 x 2) = 8140(1.0646) = $8,666

Total amount she had = $9,746 + $8,666 = $18,412

5 0

2 years ago

Read 2 more answers

In each of Problems 5 through 10, verify that each given function is a solution of the differential equation.

WARRIOR [948]

Answer:

For First Solution: $y_1(t)=e^t$

$y_1(t)=e^t$ is the solution of equation y''-y=0.

For 2nd Solution: $y_2(t)=cosht$

$y_2(t)=cosht$ is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: $y_1(t)=e^t$

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

$y_1(t)=e^t$

First order derivative:

$y'_1(t)=e^t$

2nd order Derivative:

$y''_1(t)=e^t$

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence $y_1(t)=e^t$ is the solution of equation y''-y=0.

For 2nd Solution:

$y_2(t)=cosht$

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

$y_2(t)=cosht$

First order derivative:

$y'_2(t)=sinht$

2nd order Derivative:

$y''_2(t)=cosht$

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence $y_2(t)=cosht$ is the solution of equation y''-y=0.

3 0

2 years ago

Nikki drew a rectangle with a perimeter of 18 units on a coordinate grid. Two of the vertices were (4, –3) and (–1, –3). What co

Talja [164]

Answer:

There are two possible solutions for the other two vertices of the rectangle:

(i) (4, 1), (-1, 1), (ii) (4, -7), (-1, -7)

Step-by-step explanation:

Geometrically speaking, the perimeter of a rectangle ( $p$ ) is:

$p = 2\cdot b + 2\cdot h$ (1)

Where:

$b$ - Base of the rectangle.

$h$ - Height of the rectangle.

Let suppose that the base of the rectangle is the line segment between (4, -3) and (-1, -3). The length of the base is calculated by Pythagorean Theorem:

$b = \sqrt{[(-1)-4]^{2}+[(-3)-(-3)]^{2}}$