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Travka [436]
3 years ago
14

(4 x 10e5) (2 x 10e3)

Mathematics
1 answer:
wlad13 [49]3 years ago
5 0
400,000
because if you do 10 with the exponent of 5 you'll get 100,000 multiply by 4 = 400,00
1000x2 = 2000
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Rebecca wants to prove that if the diagonals in parallelogram are perpendicular, then it is a rhombus. Select the appropriate re
VladimirAG [237]

Answer:

The answer is C

Step-by-step explanation:

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7 0
3 years ago
Jessie works at a car manufacturing plant. One day she installed a total of 46 axles, 2 in each car she worked on. She wants to
shtirl [24]

Answer:

<h2>             23 cars</h2>

Step-by-step explanation:

total number of cars:    x

the total number of axles:    46

and the number of axles  installed per car:   2

2 axles in each car, then total instaled axles:  2•x

2•x = 46

x = 46÷2

x = 23

5 0
3 years ago
What is the measure of angle ABD in trapezoid ABCD?<br><br> 24°<br> 40°<br> 64°<br> 92°
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3 0
4 years ago
Given: sin ∅= 4/5 and cos x = -5/13 ; evaluate the following expression.<br><br> tan( ∅ - x )
Pavlova-9 [17]

By definition of tangent,

tan(θ - x) = sin(θ - x) / cos(θ - x)

Expand the sine and cosine terms using the angle sum identities,

sin(x ± y) = sin(x) cos(y) ± cos(x) sin(y)

cos(x ± y) = cos(x) cos(y) ∓ sin(x) sin(y)

from which we get

tan(θ - x) = (sin(θ) cos(x) - cos(θ) sin(x)) / (cos(θ) cos(x) + sin(θ) sin(x))

Also recall the Pythagorean identity,

cos²(x) + sin²(x) = 1

from which we have two possible values for each of cos(θ) and sin(x):

cos(θ) = ± √(1 - sin²(θ)) = ± 3/5

sin(x) = ± √(1 - cos²(x)) = ± 12/13

Since there are 2 choices each for cos(θ) and sin(x), we'll have 4 possible values of tan(θ - x) :

• cos(θ) = 3/5, sin(x) = 12/13 :

tan(θ - x) = -56/33

• cos(θ) = -3/5, sin(x) = 12/13 :

tan(θ - x) = 16/63

• cos(θ) = 3/5, sin(x) = -12/13 :

tan(θ - x) = -16/63

• cos(θ) = -3/5, sin(x) = -12/13 :

tan(θ - x) = 56/33

7 0
2 years ago
Read 2 more answers
Simplifying Nonperfect Roots
Bond [772]
Short Answer:Choice C [Third one down]
Remark

Start with the number so I can talk about the basic principle at work. The way to work it is to factor 162 into prime factors and hope there are at least 3 that are the same. 
162 = 2 * 81
162 = 2 * 3 * 3 * 3 * 3

Now When you take the cube root of that, you get \sqrt[3]{2*2*2*2*3}
Here's the rule for a cube root. For every 3 prime factors under the cuberoot sign, you take out one and throw the other two away. So for cube root of 81,
you would factor it as ∛(81) = ∛(3 * 3 * 3 * 3) = 3∛3.

So out of four 3s under the cube root sign,  you have 1 outside the root sign and one inside the root sign. 2 of the four 3s have been thrown away.

Continuation.
X first
You want 2 xs outside the root sign
∛(x * x * x * x * x *x ) = (x * x)  You have thrown away 2 xs for every x outside the cube root sign
C = 6 There are no left overs.

Y second
For the ys, you need 1 outside and 2 inside the root sign. that's because you need 5 altogether.
∛(y * y * y * y * y) = y ∛y^2

Answer 
c = 6; d = 2
Choice C <<<<<< answer 
3 0
3 years ago
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