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3 years ago

For which equation would x=5 be a solution?

Mathematics

1 answer:

kari74 [83]3 years ago

8 0

03x+6=21

3(5) = 15

15+6=21

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Solve the equation using the Zero-Product Property (X-9)(x+7)=0

aleksandr82 [10.1K]

(x-9)(x+7)=0
x-9=0 or x+7=0
x=9 or x=-7

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4 0

3 years ago

The volume of each cone round the answer to decimal places

snow_tiger [21]

Answer:

1. 322.01

Step-by-step explanation:

Formula: v=pi X r^2(h/3)

4 0

3 years ago

The product 3 and a number x is 2/3. What is the value of x? A. X=7/3 B. x=2/9 C. x=2 D. X= 2 1/3

BabaBlast [244]

Answer:

the answer is C.

Step-by-step explanation:

3(2/3)

=3×2/3

=2×3/3

7 0

4 years ago

HELP PLSSSSS Triangle ABC is an equilateral triangle. The angle bisectors and the perpendicular bisectors meet at D in such a wa

vazorg [7]

CE = CD + DE = 2DE + DE = 3DE = 8.7

DE = 8.7/3 = 2.9
CD = 2DE = 5.8

For the inscribed circle, where center is point D. The radius is length DE.
This is because the circle will touch the 3 points where the bisectors intersect with sides of triangle.
radius = 2.9

For circumscribed circle, the center will also be point D. The radius is CD.
The circle will touch 3 points of triangle, A,B,C.
radius = 5.8

6 0

3 years ago

Please help!!!

Leto [7]

Answer:

- The scientist can use these two measurements to calculate the distance between the Sun and the shooting star by applying one of the trigonometric functions: Cosine of an angle.

- The scientist can substitute these measurements into $cos\alpha=\frac{adjacent}{hypotenuse}$ and solve for the distance between the Sun and the shooting star (which would be the hypotenuse of the righ triangle).

Step-by-step explanation:

You can observe in the figure attached that "AC" is the distance between the Sun and the shooting star.

Knowing the distance between the Earth and the Sun "y" and the angle x°, the scientist can use only these two measurements to calculate the distance between the Sun and the shooting star by applying one of the trigonometric functions: Cosine of an angle.

This is:

$cos\alpha=\frac{adjacent}{hypotenuse}$

In this case:

$\alpha=x\°\\\\adjacent=BC=y\\\\hypotenuse=AC$

Therefore, the scientist can substitute these measurements into $cos\alpha=\frac{adjacent}{hypotenuse}$ , and solve for the distance between the Sun and the shooting star "AC":

$cos(x\°)=\frac{y}{AC}$

$AC=\frac{y}{cos(x\°)}$

3 0

4 years ago

Read 2 more answers