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Masteriza [31]
3 years ago
7

2 Points Which of the following is (are) the solution) to (x+8)=1?

Mathematics
1 answer:
Brums [2.3K]3 years ago
7 0

Answer:

x=-7

Step-by-step explanation:

x+8=1

- Subtract 8 from both sides.

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Laura creates a rectangular prism with wooden cubes. Each cube has an edge length of 3/4 inch. she uses a total of 240 cubes. th
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Answer:00.24

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Just add the decimal form and multiply the 240 with the decimal of the 3/4

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A retailer purchased a camera for rs.1560 and sold it at 5% loss find the loss amount​
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Loss is Rs. 78

Step-by-step explanation:

Solution:

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L%= L/CP ×100%

or,. 5 =L/1560×100

or,. 5×1560/100 =L

or,. L= Rs.78

Therefore, Loss is Rs.78

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What is the midline equation for the function h(x)=−3cos(πx+2)−6?
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6 0
2 years ago
A vector with magnitude 9 points in a direction 190 degrees counterclockwise from the positive x axis. Write in component form
Over [174]

Answer:

\vec{v}= \text{ or } \approx

Step-by-step explanation:

Component form of a vector is given by \vec{v}=, where i represents change in x-value and j represents change in y-value. The magnitude of a vector is correlated the Pythagorean Theorem. For vector \vec{v}=, the magnitude is ||v||=\sqrt{i^2+j^2.

190 degrees counterclockwise from the positive x-axis is 10 degrees below the negative x-axis. We can then draw a right triangle 10 degrees below the horizontal with one leg being i, one leg being j, and the hypotenuse of the triangle being the magnitude of the vector, which is given as 9.

In any right triangle, the sine/sin of an angle is equal to its opposite side divided by the hypotenuse, or longest side, of the triangle.

Therefore, we have:

\sin 10^{\circ}=\frac{j}{9},\\j=9\sin 10^{\circ}

To find the other leg, i, we can also use basic trigonometry for a right triangle. In right triangles only, the cosine/cos of an angle is equal to its adjacent side divided by the hypotenuse of the triangle. We get:

\cos 10^{\circ}=\frac{i}{9},\\i=9\cos 10^{\circ}

Verify that (9\sin 10^{\circ})^2+(9\cos 10^{\circ})^2=9^2\:\checkmark

Therefore, the component form of this vector is \vec{v}=\boxed{}\approx \boxed{}

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