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3 years ago

A number devised by six is 306. What is the number?

Mathematics

1 answer:

QveST [7]3 years ago

8 0

306 × 6 = the number

306 × 6 = 1836

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What is the perimeter of the quadrilateral?

olasank [31]

Answer:

24.6

Step-by-step explanation:

you multiply them all by two then add them together. :P

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3 years ago

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A 16- ounce bottle of fresh water is $1.76. A 20-ounce bottle of spring water is $2.40. Which statement about the unit price is

PSYCHO15rus [73]

To find the unit price of each item you want to divide:

Fresh water:

$1.76 / 16 = $0.11

Spring water:

$2.40 / 20 = $0.12

So the statement that is true is B) Fresh water has a lower unit price of $0.11/ounce.

Hope this Helps!!

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3 years ago

1. 5 cups to 8 cups 2. 150 pounds to 135 pounds

STatiana [176]

The answer would be 60 to 30% i hope this helps<33

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3 years ago

∠A and ∠B are supplementary. m∠A = (6x + 5), m∠B = (2x + 15). Find the measure of each angle.

Anastaziya [24]

Answer:

125 and 55

Step-by-step explanation:

180=8x+20

160=8x

x=20

6(20)+5 2(20)+15=

120+5=125 degrees 40+15=55 degrees

4 0

4 years ago

Find the volume and area for the objects shown and answer Question

klio [65]

Step-by-step explanation:

You must write formulas regarding the volume and surface area of the given solids.

$\bold{\#1\ Rectangular\ prism:}\\\\V=lwh\\SA=2lw+2lh+2wh=2(lw+lh+wh)\\\\\bold{\#2\ Cylinder:}\\\\V=\pi r^2h\\SA=2\pi r^2+2\pi rh=2\pir(r+h)\\\\\bold{\#3\ Sphere:}\\\\V=\dfrac{4}{3}\pi r^3\\SA=4\pi r^2$

$\bold{\#4\ Cone:}\\\\V=\dfrac{1}{3}\pi r^2h\\\\\text{we need calculate the length of a slant length}\ l\\\text{use the Pythagorean theorem:}\\\\l^2=r^2+h^2\to l=\sqrt{r^2+h^2}\\\\SA=\pi r^2+\pi rl=\pi r^2+\pi r\sqrt{r^2+h^2}=\pi r(r+\sqrt{r^2+h^2})\\\\\bold{\#5\ Rectangular\ Pyramid:}\\\\V=\dfrac{1}{3}lwh\\\\$

$\\\text{we need to calculate the height of two different side walls}\ h_1\ \text{and}\ h_2\\\text{use the Pythagorean theorem:}\\\\h_1^2=\left(\dfrac{l}{2}\right)^2+h^2\to h_1=\sqrt{\left(\dfrac{l}{2}\right)^2+h^2}=\sqrt{\dfrac{l^2}{4}+h^2}=\sqrt{\dfrac{l^2}{4}+\dfrac{4h^2}{4}}\\\\h_1=\sqrt{\dfrac{l^2+4h^2}{4}}=\dfrac{\sqrt{l^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{l^2+4h^2}}{2}$

$\\\\h_2^2=\left(\dfrac{w}{2}\right)^2+h^2\to h_2=\sqrt{\left(\dfrac{w}{2}\right)^2+h^2}=\sqrt{\dfrac{w^2}{4}+h^2}=\sqrt{\dfrac{w^2}{4}+\dfrac{4h^2}{4}}\\\\h_2=\sqrt{\dfrac{w^2+4h^2}{4}}=\dfrac{\sqrt{w^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{w^2+4h^2}}{2}$

$SA=lw+2\cdot\dfrac{lh_1}{2}+2\cdot\dfrac{wh_2}{2}\\\\SA=lw+2\!\!\!\!\diagup\cdot\dfrac{l\cdot\frac{\sqrt{l^2+4h^2}}{2}}{2\!\!\!\!\diagup}+2\!\!\!\!\diagup\cdot\dfrac{w\cdot\frac{\sqrt{w^2+4h^2}}{2}}{2\!\!\!\!\diagup}\\\\SA=lw+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw}{2}+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw+l\sqrt{l^2+4h^2}+w\sqrt{w^2+4h^2}}{2}$

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3 years ago