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Citrus2011 [14]
4 years ago
5

How many multiples of 7 are there between 100 and 1000.

Mathematics
2 answers:
vivado [14]4 years ago
8 0
So after 100 its 105,112,119...........994
x=7(multiples of what we're finding), y=105(where we start from after 100), a multiplied by z=994
we have to find z

994=105+(z-1)7
994-105=7z-7
889+7=7z
896=7z
z= 128
Alex Ar [27]4 years ago
8 0
105,112,119,126,133,140,147,154,161,168,175,182,189,196,203,210,217,224,231,238,245,252,259,266,273,280,287,294,301,308,315,322,329,336,343, 350, 357, 364, 371, 378, 385, 392, 399, 406, 413, 420, 427, 434, 441, 448, 455, 462, 469, 476, 483, 490, 497, 504, 511, 518, 525, 532, 539, 546, 553, 560, 567, 574, 581, 588, 595, 602, 609, 616, 623, 630, 637, 644, 651, 658, 665, 672, 679, 686, 693, 700, 707, 714, 721, 728, 735, 742, 749, 756, 763, 770, 777, 784, 791, 798, 805, 812, 819, 826, 833, 840, 847, 854, 861, 868, 875, 882, 889, 896, 903, 910, 917, 924, 931, 938, 945, 952, 959, 966, 973, 980, 987, 994

Read more: http://www.mathwarehouse.com/arithmetic/multiple-calculator.php?#ixzz5C6UCeW5c
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If a ball is thrown into the air with a velocity of 34 ft/s, its height (in feet) after t seconds is given by y = 34t − 16t2. Fi
Finger [1]

<u>ANSWER: </u>

If a ball is thrown into the air with a velocity of 34 feet per second, then velocity of the ball after 1 second is 2 feet per second

<u>SOLUTION: </u>

Given, a ball is thrown into the air with a velocity of 34 feet per second

Initial velocity (u) = 34 feet per second

And also given a relation between displacement and time = \mathrm{y}=34 \mathrm{t}-16 \mathrm{t}^{2} --- eqn 1

We need to find the velocity when t = 1 ; v = ?

We know that, v = u + at and \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}

where v is instantaneous velocity and u is initial velocity

a is acceleration

t is time interval  

s is displacement

using the displacement and time relation eqn (1) we get

Now, when t = 1, displacement s = 34(1) – 16(1)

\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}=34-16

34 \times 1+\frac{1}{2} \times a \times 1^{2}=18

34+\frac{a}{2}=18

\begin{array}{l}{\frac{a}{2}=18-34} \\\\ {\frac{a}{2}=-16} \\ {a=-16 \times 2} \\ {a=-32}\end{array}

here, -ve sign indicates that object is in deceleration . so acceleration is -32 ft/s

now put a value in v = u + at

v = 34 + (-32)(1)

v = 34 – 32

v = 2 ft/s

Hence, velocity of the ball after 1 second is 2 ft/s

6 0
3 years ago
What is an equation of the line that passes through the point (-4,-6) and is perpendicular to the line 4x+5y=25
elixir [45]

Answer:

y =  \frac{5}{4} x - 1

Step-by-step explanation:

The equation of a line is usually written in the form of y=mx+c, where m is its gradient and c is its y-intercept.

First rewrite the equation of the given line in the form of y=mx +c.

4x+5y=25

5y= -4x +25

y =  -  \frac{4}{5} x + 5

The gradient of the given line is -  \frac{4}{5}

The product of the gradient of perpendicular lines is -1.

( -  \frac{4}{5} )(gradient \: of \: line) =  - 1 \\ gradient \: of \: line =  - 1  \div ( -  \frac{4}{5} ) \\ gradient \: of \: line =  \frac{5}{4}

Thus, m= \frac{5}{4}

y =  \frac{5}{4} x + c

Substitute a coordinate to find c.

When x= -4, y= -6,

- 6 =  \frac{5}{4} ( - 4) + c \\  - 6 =  - 5 + c \\ c = 5 - 6 \\ c =  - 1

Hence, the equation of the line is

y =  \frac{5}{4} x - 1

7 0
3 years ago
F is the function f(×)= 2x+5<br> Find f(3)
tatiyna

Answer:

f(3)=11

Step-by-step explanation:

f(x) = 2x + 5

to find f(3) we simply plug in 3 where ever x is

So f(3) = 2(3) + 5

==> multiply 2 and 3

f(3) = 6 + 5

==> add 6 and 6

f(3) = 11

5 0
2 years ago
Help meeeeee!!! please please please lol 1- 11
attashe74 [19]

Answer:

1. a = -9

2.p=4

3.r=7

4. a = -14/9

5.  k = -2

6.y=3

7.k=-12

8. x = -12

9. n = 0

10. x=0

11.x=4

8 0
3 years ago
Tiny's Tattoo Parlor sold 389 tattoos this month. 43 of those tattoos were arm tattoos. Based on this data, what is a reasonable
Dmitry [639]

389 - 43 = 346
346 are not arm tattoos

Probability of a customer not getting an arm tattoo =
\frac{346}{389}
Answer: 346/389

Hope this helps.
8 0
3 years ago
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