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Snezhnost [94]
3 years ago
6

This is due today pls help!!!!!!!

Mathematics
2 answers:
yawa3891 [41]3 years ago
8 0

Answer:

x = 3.5

Step-by-step explanation:

Each value is multiplied by 1.5 so 7 times 1.5 is 10.5

10.5/3 = 3.5

KATRIN_1 [288]3 years ago
8 0

Answer:

1.5

Step-by-step explanation:

To find the value of x that makes triangle DEF similar to XYZ, we have to find the scale. To do this find 2 similar sides. I chose 7.5 from XYZ and 5 from DEF. Divide 7.5 by 5 to get 1.5. 1.5 is our scale so this means we will insert 1.5 for x and get 4.5 as our answer.

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This is for algebra 2 and I got stuck on this anyone know this?
AnnZ [28]
You just simplify the fraction (so, 8/6 which is 4/3)

So your comes 4(square root of 35)/ 3 (square root of 15)
5 0
3 years ago
PLS HELP ME WITH THIS <br> will be reporting answers that don't actually help lol
Dmitry [639]

Answer:

x=3/radical2,-3/radical2,3/4

Step-by-step explanation:

By factoring you get

4x(2x^2-9)-3(2x^2-9)=(2x^2-9)(4x-3)=0

x^2=9/2, x=3/4

x=3radical2/2, -3radical2/2 , 3/4

5 0
3 years ago
Wjat are the x intercepts if y=x squared minus 100​
Effectus [21]

Answer:

x = ± 10

Step-by-step explanation:

Given

y = x² - 100 ← a difference of squares , that is

y = x² - 10² = (x - 10)(x + 10)

To find the x- intercepts let y = 0 , that is

(x - 10)(x + 10) = 0

Equate each factor to zero and solve for x

x - 10 = 0 ⇒ x = 10

x + 10 = 0 ⇒ x = - 10

3 0
3 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
3 years ago
0.075 as a fraction
Rudiy27
.075 will be written as 75/1,000. The fraction can then be simplified to the fraction 3/40.
8 0
3 years ago
Read 2 more answers
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