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2 years ago

Find the next term of each sequence 25,17,9,1...

Mathematics

1 answer:

Karolina [17]2 years ago

7 0

Answer:

your answer in ☝️ this pic

Step-by-step explanation:

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Is this prime or can it be factored?

poizon [28]

Factor the following:
6 x^2 - 19 x + 10
Factor the quadratic 6 x^2 - 19 x + 10. The coefficient of x^2 is 6 and the constant term is 10.
The product of 6 and 10 is 60. The factors of 60 which sum to -19 are -4 and -15.
So 6 x^2 - 19 x + 10 = 6 x^2 - 15 x - 4 x + 10 = 2 x (3 x - 2) - 5 (3 x - 2):
2 x (3 x - 2) - 5 (3 x - 2)
Factor 3 x - 2 from 2 x (3 x - 2) - 5 (3 x - 2):
Answer: (3 x - 2) (2 x - 5)

3 0

3 years ago

Calculate the slope of each segment (just need a couple example problems and step by step, no need to answer all. Will mark brai

Tpy6a [65]

Answer:

12) Line segment AB

$m_{AB} = \frac{7-4}{10-4}= \frac{1}{2}$

13) Line segment CD

$m_{CD} = \frac{6-11}{15-13}=-\frac{5}{2}$

14) Line segment EF

$m_{EF} = \frac{7-11}{4-3}= -4$

15) Line segment GH

$m_{GH} = \frac{5-2}{8-6}= \frac{3}{2}$

16) Line segment IJ

$m_{IJ} = \frac{0-(-6)}{0-4}=-\frac{3}{2}$

17) Line segment KL

$m_{KL} = \frac{-2-(-3)}{5-(-2)} = \frac{1}{7}$

18) Line segment MN

$m_{MN} = \frac{5-5}{3-(-4)} = 0$

19) Line segment OP

$m_{OP} = \frac{4-(-2)}{13-13} = \frac{1}{0}$

20) Line segment QR

$m_{QR} = \frac{7-11}{-2-(-7)} = -\frac{4}{5}$

21) Line segment ST

$m_{ST} = \frac{-2-(-3)}{10-9} = 1$

Step-by-step explanation:

Coordinates are now described: $A(x,y) = (4,4)$ , $B(x,y) = (10, 7)$ , $I(x,y) = (-6, 4)$ , $J(x,y) = (0,0)$ , $Q(x,y) = (-7,11)$ , $R(x,y) = (-2,7)$ , $C(x,y) = (13,11)$ , $D(x,y) = (15, 6)$ , $K(x,y) = (-2, -3)$ , $L(x,y) = (5,-2)$ , $S(x,y) = (9,-3)$ , $T(x,y) = (10, -2)$ , $E(x,y) = (3, 11)$ , $F(x,y) = (4,7)$ , $M(x,y) = (-4,5)$ , $N(x,y) = (3, 5)$ , $G(x,y) = (6,2)$ , $H(x,y) = (8,5)$ , $O(x,y) = (13,-2)$ , $P(x,y) = (13,4)$

The slope of each line segment is calculated by applying definition of secant line:

12) Line segment AB

$m_{AB} = \frac{7-4}{10-4}= \frac{1}{2}$

13) Line segment CD

$m_{CD} = \frac{6-11}{15-13}=-\frac{5}{2}$

14) Line segment EF

$m_{EF} = \frac{7-11}{4-3}= -4$

15) Line segment GH

$m_{GH} = \frac{5-2}{8-6}= \frac{3}{2}$

16) Line segment IJ

$m_{IJ} = \frac{0-(-6)}{0-4}=-\frac{3}{2}$

17) Line segment KL

$m_{KL} = \frac{-2-(-3)}{5-(-2)} = \frac{1}{7}$

18) Line segment MN

$m_{MN} = \frac{5-5}{3-(-4)} = 0$

19) Line segment OP

$m_{OP} = \frac{4-(-2)}{13-13} = \frac{1}{0}$

20) Line segment QR

$m_{QR} = \frac{7-11}{-2-(-7)} = -\frac{4}{5}$

21) Line segment ST

$m_{ST} = \frac{-2-(-3)}{10-9} = 1$

8 0

2 years ago

Someone help me please

Mariana [72]

Answer: I think 10

Step-by-step explanation: cuz area is length x height and basically you just do square root of 100 Bc thags a square so the sides should be equal

8 0

3 years ago

Select all the shapes below that have a constant cross-section between a

meriva

The shapes below that have a constant cross-section between a pair of faces are cylinder, cube, sphere, hexagonal, and prism.

<h3>What is the definition of geometry?</h3>

It is concerned with the geometry, region, and density of various 2D and 3D shapes.

A cylinder is a closed solid with two parallel circular bases and a curving surface connecting them.

The shapes have a constant cross-section between a pair of faces like a cylinder.

Hence,cylinder, cube, sphere, hexagonal, and prism are the forms below that have a constant cross-section between a pair of faces.

To learn more about geometry, refer to brainly.com/question/7558603.

#SPJ1

5 0

2 years ago

A recipe uses 1/4 cup of sugar for every 1/2 teaspoon of cinnamon . What is the unit rate in teaspoons per cup?

Romashka-Z-Leto [24]

Answer:

Step-by-step explanation:

1/4 cup of sugar = 1/2 tea spoon. (1)

Multiplying both sides by 4, we have

1/4 * 4 cup of sugar = 1/2 * 4 tea spoon. (2)

1 cup of sugar = 2 tea spoon

( Required answer)

5 0

3 years ago

Read 2 more answers