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2 years ago

Priya has a recipe for banana bread. She uses

Mathematics

2 answers:

Artyom0805 [142]2 years ago

4 0

I think it’s b hope I’m right

Illusion [34]2 years ago

3 0

Answer:

i think its B, Hope u ace it sorry if its wrong i truly tried my best

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A figure is composed of a semicircle and a right triangle. Determine the area of the shaded region. Use 3.14 for π and round to

Aliun [14]

Answer:

$A=9.5\ ft^2$

Step-by-step explanation:

we know that

The area of the shaded region is equal to the area of triangle plus the area of semicircle

$A=\frac{1}{2}bh+\frac{1}{2}\pi r^{2}$

we have

$b=4\ ft$

Find the height of the right triangle applying the Pythagorean Theorem

$5^2=h^2+4^2$

$h^2=9\\h=3\ ft$

The radius of the semicircle is half the height of triangle

$r=3/2=1.5\ ft$

substitute in the formula

$A=\frac{1}{2}(4)(3)+\frac{1}{2}(3.14)(1.5)^{2}$

$A=6+3.5=9.5\ ft^2$

5 0

2 years ago

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Help me and this question please please

disa [49]

Answer:

the things wont load whats the question?

Step-by-step explanation:

8 0

2 years ago

A guy wire 30 ft long supports an antenna at a point that is 24 ft above the base of the antenna.

GalinKa [24]

Use Pythagorean Theorem
c2 = a2 + b2
152 = a2 + 12
a2 = 152 - 122
=225 - 144 = 81
a = 9 ft
Distance from base of antenna = 9ft

4 0

2 years ago

Find the slope of line a and line b from two points

Degger [83]

Answer: Line A has a slope of 3/4 and Line B has a slope of 5/6

Step-by-step explanation: Look at the attachment

5 0

2 years ago

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Hello people ~

Luden [163]

Cone details:

height: h cm

radius: r cm

Sphere details:

radius: 10 cm

================

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

Using Pythagoras Theorem

(a)

TO² + TU² = OU²

(h-10)² + r² = 10² [insert values]

r² = 10² - (h-10)² [change sides]

r² = 100 - (h² -20h + 100) [expand]

r² = 100 - h² + 20h -100 [simplify]

r² = 20h - h² [shown]

r = √20h - h² ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (\sqrt{20h - h^2})^2 \ ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20h - h^2) (h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20 - h) (h) ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)$

To find maximum/minimum, we have to find first derivative.

(c)

First derivative

$\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )$

apply chain rule

$\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}$

Equate the first derivative to zero, that is V'(x) = 0

$\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0$

$\Longrightarrow \sf 40h-3h^2=0$

$\Longrightarrow \sf h(40-3h)=0$

$\Longrightarrow \sf h=0, \ 40-3h=0$

$\Longrightarrow \sf h=0,\:h=\dfrac{40}{3}$

maximum volume: when h = 40/3

$\sf \Longrightarrow max= \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )$

$\sf \Longrightarrow maximum= 1241.123 \ cm^3$

minimum volume: when h = 0

$\sf \Longrightarrow min= \dfrac{1}{3} \pi (0)^2(20-0)$

$\sf \Longrightarrow minimum=0 \ cm^3$

6 0

2 years ago

Read 2 more answers