Please help! Number 5 please!

Mathematics

1 answer:

Vaselesa [24]3 years ago

8 0

Answer:

No.

Step-by-step explanation:

The first gym, by adding the first is 26.05 dollars and the other one would cost you 17.05 dollars.

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(15.) A toy is in the shape of a cone mounted on a hemisphere of same base

lisov135 [29]

Given:

Volume of toy=231 cm³

Diameter of a hemisphere= 7 cm

cone and hemisphere have equal radius.

radius of hemisphere = radius of cone = 3.5 cm

Height of hemisphere= radius of hemisphere= 3.5 cm

Let H be the height of toy

H= height of cone+ height of hemisphere

H = h + r , ( h = height of cone)

H = h + 3.5

volume of toy = volume of cone + volume of hemisphere

Volume of toy= 1/3πr²h + 2/3πr³

Volume of toy=πr²/3(h+2r)

231 = (22/7)×(3.5)² ×(1/3)(h+2×3.5)

231×3 =( 22×3.5×3.5)/7 (h+7)

h+7 = (231×3×7)/(3.5×3.5×22)

h+7 = (3×3)/(.5×.5×2)

h+7= 900 /50= 90/5= 18

h+7= 18

h=18-7

h= 11 cm

Height of toy = h+r

Height of toy = 11+3.5= 14.5

Height of toy =14.5 cm

Hence, the height of the toy = 13.5 cm

3 0

3 years ago

Tell the shape, center, and spread of this data set.

lions [1.4K]

Answer:

can I have options? or is there not any?

8 0

3 years ago

In the right ΔABC, AN is the altitude to the hypotenuse. Find BN, AN, and AC, if AB=2√5 in and NC= 1 in.

kipiarov [429]

Answer:

$BN=4\ \text{in}$

$AN=2\ \text{in}$

$AC=\sqrt{5}\ \text{in}$

Step-by-step explanation:

Let $BN=x$

$BC=c$

$NC=y=1\ \text{in}$

$AB=a=2\sqrt{5}\ \text{in}$

$AC=b$

We have the relation

$\dfrac{BC}{AB}=\dfrac{AB}{BN}\\\Rightarrow \dfrac{c}{a}=\dfrac{a}{x}\\\Rightarrow c=\dfrac{a^2}{x}\\\Rightarrow x+1=\dfrac{a^2}{x}\\\Rightarrow x^2+x=20\\\Rightarrow x^2+x-20=0\\\Rightarrow x=\frac{-1\pm \sqrt{1^2-4\times 1\times \left(-20\right)}}{2\times 1}\\\Rightarrow x=4,-5$