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3 years ago

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The lengths of the diagonals of a rhombus are 2 in. and 5 in. Find the measures of the angles of the rhombus to the nearest degr

ee.

1 answer:

cricket20 [7]3 years ago

6 0

Asked and answered elsewhere.
brainly.com/question/8709463

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A spacecraft is traveling with a velocity of v0x = 5320 m/s along the +x direction. Two engines are turned on for a time of 739

OlgaM077 [116]

Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

For a constant acceleration: $v_{f}=v_{0}+at$ , where $v_{f}$ is the final velocity in a direction after the acceleration is applied, $v_{0}$ is the initial velocity in that direction before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
<em>Then for the x direction</em> it is known that the initial velocity is $v_{0x} =$ 5320 m/s, the acceleration (the applied by the engine) in x direction is $a_{x}$ 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the is 739 s. Then: $v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}$
In the same fashion, <em>for the y direction</em>, the initial velocity is $v_{0y} =$ 0 m/s, the acceleration in y direction is $a_{y}$ 7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$

8 0

3 years ago

HELP!! 70 points will be given!<br> Picture included

kkurt [141]

Answer: a) -13/16

Step-by-step explanation: Start by setting equations equal and rearrange X^3 - x^2 + 1 = 0. Visual inspection of graph shows x between -1 and -1/2. Start with x = 3/4 plug in and calculate: just a little too small. Try going halfway towards -1: x =-7/8 Plug in and the answer is very far from 0. Go halfway back towards -3/4: -13/16 and the equality is very close.

8 0

4 years ago

Make g the subject of this relation

zhannawk [14.2K]

Answer:

see explanation

Step-by-step explanation:

Simplify the radical

$\frac{U}{\frac{1}{f}+\frac{1}{g} }$

= $\frac{U}{\frac{g+f}{fg} }$

= $\frac{Ufg}{g+f}$

Square both sides

T² = $\frac{Ufg}{g+f}$ ( multiply both sides by (g + f) )

T²(g + f) = Ufg ( distribute left side )

T²g + T²f = Ufg ← subtract Ufg from both sides

T²g - Ufg + T²f = 0 ← subtract T²f from both sides

T²g - Ufg = - T²f ← factor out g from each term on the left side

g(T² - Uf) = - T²f ← divide both sides by (T² - Uf)

g = - $\frac{T^2f}{T^2-Uf}$ = $\frac{T^2f}{Uf-T^2}$

6 0

3 years ago

A direct relationship occurs when

amid [387]

A direct relationship requires a change in the same direction of both variables, answer c.

4 0

3 years ago

Twice the product of 3and a number exceeds the sum of the number and 80 by 10.what is the number?

Hatshy [7]

First, create an equation with the criteria given.

If we let x be your number,
$2(3x) = 80 + x + 10$
( $2(3x)$ is twice the product of 3 and x, and if it exceeds $x+80$ by 10, then we add 10 to $x+80$ to make the two sides of the equation equal.)

Now, solve for $x$ .

$6x=90+x \\ 5x=90 \\ x=18$

6 0

3 years ago