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1 year ago

Find in degrees. 11 B 6 [?]° Round to the nearest hundredth. Enter

Mathematics

1 answer:

avanturin [10]1 year ago

5 0

Answer:

$\beta = 28.610$

Step-by-step explanation:

in this case we have a right angled triangle, whenever you see this triangle think of the TRIG RATIOS * sin, cos & tan* but if it was another triangle besides right angled triangle you use SINE RULE OR COSINE RULE

SOLUTION

$\tan( \beta ) = \frac{opposite}{adjecent} \\ \tan( \beta ) = \frac{6}{11} \\ \beta = tan(inverse)( \frac{6}{11} ) \\ \beta = 28.61045$

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The figure in Quadrant III of the coordinate plane below is a transformation of the figure in quadrant II.

BARSIC [14]

Answer: 2nd option, 3rd option, and the last option are all correct

5 0

3 years ago

Order these numbers from least to greatest. 6.64 , 467 , 44 , −45

lutik1710 [3]

-45, 6.64, 44, 467
Least Greatest

5 0

3 years ago

Allison has three containers with 25 crayons in each. She also has four boxes of markers with 12 markers in each box. She gets 1

Advocard [28]

25x3=75
4x12=48
48-10=38

75 crayons and 38 markers

3 0

3 years ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$

Since time can only be positive we reject the $t_{2}$ solution and we keep that Ben's book took $t=4.3276$ seconds to reach the ground.

Similarly solving for Josh we obtain

$t_{1}=3.6794sec\\t_{2}=-0.6794sec$

Thus again we reject the negative and keep the positive solution, so Josh's book took $t=3.6794$ seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

$t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482$

So to answer the original question:

Whose textbook reaches the ground first and by how many seconds?

Josh's textbook reached the ground first
Josh's textbook reached the ground first by a difference of $t=0.6482$

3 0

2 years ago

URGENT!!LAST QUESTION!!!! PLS JUST TAKE A LOOK! EASY BUT I AM DUMB!!!!!! WILL GIVE BRANLIEST!!

Luden [163]

Answer:

x = 28

Step-by-step explanation:

If the quadrilaterals are similar, there is a proportionality among their sides:

The top side in the large figure (70) is to the top side in the small figure (10) in the same ratio as the left side (x) in the large figure is to the left side in the small figure (4). This in math terms is written as:

$\frac{70}{10} =\frac{x}{4}$

We can then solve for the unknown "x" by multiplying both sides by 4:

$\frac{70}{10} =\frac{x}{4}\\\frac{70\,*\,4}{10} =x\\x=28$

4 0

2 years ago