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3 years ago

10

Find the slope of the line connecting (1, 7), (10, 1)

1 answer:

kari74 [83]3 years ago

7 0

Slope = (Y2 -Y1) ÷ (X2 -X1)

Slope = (1 -7) / (10 -1)

Slope = -6 / 9 = -0.6666666666666666

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Gente me ajudem prfvv em matemática

Tasya [4]

Answer:

Step-by-step explanation:

Trace a linha de 8cm, daí você vai dividir ela no meio, ou seja, no 4. Aí é só contar 2 cm para cada lado, que vai ser o 1, as pontas vão ser o 2. Tendeu? Daí é só marcar os pontos no -3/7(~-2,3), 1,6, 7/5(1,4), -1 e 0

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2 years ago

Need help, I did a similar problem and got it wrong soo...

Karo-lina-s [1.5K]

Percent change = (new number - old number)/(old number) * 100

percent change = (54 - 48)/(48) * 100 = 6/48 * 100 = 0.125 * 100 = 12.5%

8 0

3 years ago

Read 2 more answers

A necklace found at a pawn shop is 192 grams and has a volume of 19.5 cubic centimetres. What metal is the necklace made of?

Sophie [7]

Answer:

we divide 192 by the volume 19.5

192÷19.5=9.85 so the necklace is made of bronze

8 0

3 years ago

Arrange the cones in order from lease volume to greatest volume

bearhunter [10]

Answer:

Volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let $V_{2}. V_{3}. and\ V_{4}.$ be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

$d_{1} = 20\ and\ h_{1}=12$

$d_{2} = 18\ and\ h_{2}=10$

$r_{3} = 10\ and\ h_{3}=9$

$r_{4} = 11\ and\ h_{14}=9$

Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

$V=\pi r^{2} \frac{h}{3}$ ----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for $d_{1} = 20\ and\ h_{1}=12$

$r_{1} = \frac{d_{1}}{2}$

$r_{1} = \frac{20}{2}=10\ units$

$V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}$

$V_{1}=\pi (10)^{2} \frac{12}{3}$

$V_{1}=\pi\times 100\times 4$

$V_{1}=400\pi\ units^{3}$

Similarly, for volume of the cone for $d_{2} = 18\ and\ h_{2}=10$

$r_{2} = \frac{d_{2}}{2}$

$r_{2} = \frac{18}{2}=9\ units$

$V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}$

$V_{2}=\pi (9)^{2} \frac{10}{3}$

$V_{2}=\pi\times 81\times \frac{10}{3}$

$V_{2}=\pi\times 27\times 10$

$V_{2}=270\pi\ units^{3}$

Similarly, for volume of the cone for $r_{3} = 10\ and\ h_{3}=9$

$V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}$

$V_{3}=\pi (10)^{2} \frac{9}{3}$

$V_{3}=\pi\times 100\times 3$

$V_{3}=\pi\times 300$

$V_{3}=300\pi\ units^{3}$

Similarly, for volume of the cone for $r_{4} = 11\ and\ h_{4}=9$

$V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}$

$V_{4}=\pi (11)^{2} \frac{9}{3}$

$V_{4}=\pi\times 121\times 3$

$V_{4}=\pi\times 363$

$V_{4}=363\pi\ units^{3}$

So, the volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

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3 years ago

a frog is at the bottom of a 25 foot well. each day he climbs up 3 feet, and each night he slips down 2 feet. how many days will

klemol [59]

25. he goes up by one foot (3 minus 2) each day and night, so 25.

6 0

1 year ago