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2 years ago

PLEASE HELP ME FAST QUESTION IN PHOTO WILL PICK BRAINLIEST IF CORRECT

Mathematics

1 answer:

Annette [7]2 years ago

5 0

Where is the photo? I don't see a picture

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A right triangle is given with a missing value of t. It is stated that the triangle is an acute right triangle with angles 50∘ a

sergiy2304 [10]

Answer: t=20°

Step-by-step explanation:

1. By definition, the sum of the interior angles is 180 degrees.

2. Keeping the information above on mind, you can write the following expression:

$90+50+2t=180$

3. Now, you must solve for $t$ as following:

$140+2t=180\\2t=180-140\\2t=40\\t=\frac{40}{2}\\t=20$

4. Therefore, the value of $t$ is 20°

6 0

3 years ago

How do you do this??

astra-53 [7]

welll...you have there written just one relationship...it is berween 280 and 4 hours...you have to figure out speed:

$280 \div 4 = 70$

it is speed per hour

then you just connect time and speed:

140 - 2
280 - 4
350 - 5
420 - 6
560 - 8
700 - 10

6 0

2 years ago

Alyssa was given a gift card for a coffee shop each morning Alyssa uses the gift card to buy one cup of coffee let a represent t

raketka [301]

Answer:20.00

Step-by-step explanation: next time include the table but i managed to find it anyway

5 0

7 months ago

Through: (3,3), parallel to y=8/3x+5

pshichka [43]

Answer:

(0,5)

Step-by-step explanation:

5 0

2 years ago

A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

qaws [65]

The projectile's horizontal and vertical positions at time $t$ are given by

$x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t$

$y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2$

where $g=9.8\dfrac{\rm m}{\mathrm s^2}$ . Solve $y=0$ for the time $t$ it takes for the projectile to reach the ground:

$30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s$

In this time, the projectile will have traveled horizontally a distance of

$x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m$

The projectile's horizontal and vertical velocities are given by

$v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ$

$v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt$

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .

7 0

3 years ago