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1 year ago

Find an equivalent ratio in simplest terms: 63:49

Mathematics

2 answers:

Fudgin [204]1 year ago

4 0

Answer:Is 97.

Step-by-step explanation:

Ede4ka [16]1 year ago

3 0

I think it’s 9:7 because the GCF of 63 and 49 is 7. So, you would divide both numbers by 7 to get 9:7.

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Plsssssss help -1 3/10 + 2/5

vitfil [10]

Answer:

Step-by-step explanation:

-9/10 or -0.9

8 0

2 years ago

Y = 4x+1 y=x^2 + 2x - 2 A.6,2 B. 2,6 C. -6,2 D. -6,-2

natta225 [31]

Answer:

The solution is (3,13) and (-1,-3). So none of the mentioned options is correct.

Step-by-step explanation:

Given that

$y = 4x + 1 (i)$

$y = x^2 + 2x -2 (ii)$

Now, by susbstituting the value of 'y' from equation i to equation ii, we get

$4x+1=x^2 + 2x -2$

$0=x^2 + 2x -2-4x-1$

$0=x^2 + (2x -4x) +(-2-1)$

$0=x^2 + (-2x) +(-3)$

$0=x^2 -2x -3$

$x^2 -2x -3 = 0 (iii)$

Now by factorization, equation iii can be written as

$x^2 -3x +x -3 = 0$

$x(x -3) + 1(x -3) = 0$

$(x -3)(x +1) = 0$

x = 3 and x = -1

By putting the values of x in equation i, we get

y = 4(3) + 1

y = 12 +1

y = 13

and

y = 4(-1) + 1

y = -4 +1

y = -3

Therefore, the solution is (3,13) and (-1,-3). So none of the mentioned options is correct.

7 0

2 years ago

A spinner has 3 sections colored red, yellow, and blue. The probabilities of the spinner landing on red and yellow or 0.25 and 0

JulijaS [17]

It would be 37.5 times, which rounded up would be about 38 because blue would be .25 of the spinner, so you would expect it to land on blue 25% of the time. Hope this helps!!

5 0

2 years ago

System of quadratic equations 3 variables how to

WITCHER [35]

<span>when you are solving systems of equation with three variables we use the elimination method to make a system of two equations in two variables</span>

4 0

2 years ago

In quadratic drag problem, the deceleration is proportional to the square of velocity

Mars2501 [29]

Part A

Given that $a= \frac{dv}{dt} =-kv^2$

Then,

$\int dv= -kv^2\int dt \\ \\ \Rightarrow v(t)=-kv^2t+c$

For $v(0)=v_0$ , then

$v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0$

Thus, $v(t)=-kv(t)^2t+v_0$

For $v(t)= \frac{1}{2} v_0$ , we have

$\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0}$

Part B

Recall that from part A,

$v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a$

Now, at initial position, t = 0 and $v=v_0$ , thus we have

$x=a$

and when the velocity drops to half its value, $v= \frac{1}{2} v_0$ and $t= \frac{2}{kv_0}$

Thus,

$x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\ \\ =- \frac{1}{2k} + \frac{2}{k} +a$

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

$- \frac{1}{2k} + \frac{2}{k} +a-a \\ \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}$

7 0

2 years ago