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3 years ago

A triangle has one side that measures 5 ft, one side that measures 6 ft, and one side that measures 9 ft.

Mathematics

2 answers:

pochemuha3 years ago

5 0

Area= 14.142

Perimeter=20

Height=3.14

wlad13 [49]3 years ago

5 0

Answer:

TOTAL PERIMETER = 5 ft. + 6 ft. + 9 ft. = 20 ft.

TOTAL AREA = 5 ft. x 6 ft. = 30 ft. x 9 ft. = 270 ft.

Step-by-step explanation:

Scalene Triangle

Area = 14.14214

Perimeter = 20

Semi Perimeter = 10

Height A = 5.6569

Height B = 4.714

Height C = 3.1427

Median A = 7.2284

Median B = 6.6332

Median C = 3.2016

Angle A = 0.5513 rad, 31.5863° deg

Angle B = 0.6797 rad, 38.9425° deg

Angle C = 1.9106 rad, 109.4713° deg

Inradius = 1.4142

Circumradius = 4.7729

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While playing basketball this weekend Frank shoots an airball. The height H in feet of the ball is given by H equals -16t^2 + 32

Korvikt [17]

Answer:

Step-by-step explanation:

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You just want to find t when H=0. So, solve

-16t^2 + 32t + 8 = 0

t = 2.22

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6 0

3 years ago

20 POINTS!!!! <br> HELPPPPP THIS IS DUE TONIGHT!!!!

alexandr1967 [171]

Step-by-step explanation:

Our coordinates are (6, 0) and (0, -3).

We can use the formula (y₂-y₁)/(x₂-x₁) to find the slope-intercept form (y=mx+b)

Now, we plug the coordinates into the formula and solve.

(0-(-3))/(6-0)

=3/6

= $\frac{1}{2}$

So, the slope is $\frac{1}{2}$ .

Now, we have to find the y-intercept, which is the point where the line on the graph intersects with the y-axis.

We can find the y-intercept (which is represented by b) by plugging the information we already know into the formula.

We can use either pair of coordinates for the formula, so I'll use (6, 0).

y=mx+b

0= $\frac{1}{2}$ (6)+b

Now, we can solve.

0=3+b

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Answer:

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2 years ago

A number reduced by 15 results in 14. Which equation models this sentence?

Citrus2011 [14]

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4 0

3 years ago

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Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt

polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

$\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt$

$=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k$

The first integral is trivial since $(\tan t)'=\sec^2t$ .

The second can be done by substituting $u=t^2-1$ :

$u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C$

The third can be found by integrating by parts:

$u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t$

$\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8$

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8 0

3 years ago

Pls help it’s due tomorow, giving brainliest to the first answer with all three correct :)

aev [14]

Answer:

b im positive it B but if im wrong message me

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3 years ago

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