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3 years ago

2,617 rounded to the nearest ten

Mathematics

2 answers:

r-ruslan [8.4K]3 years ago

8 0

Answer:

2,620.

Step-by-step explanation:

thats the answer. have a nice day!

fgiga [73]3 years ago

3 0

The answer would be 2,620 because 17 is closest to 20 so you would round to the closest ten to 17

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Which number is NOT in the solution set of the inequality x > −15?

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Answer:

A = 53 degree

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What are the x- and y-intercepts of the graph of 3x - 4y = 9?

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Answer:

The X and Y intercepts and the Slope are called the line properties. We shall now graph the line -3x-4y-9 = 0 and calculate its properties. Graph of a Straight Line : Calculate the Y-Intercept : Notice that when x = 0 the value of y is 9/-4 so this line "cuts" the y axis at y=-2.25000. y-intercept = 9/-4 = -2.25000.

Step-by-step explanation:

Find the X and Y Intercepts 3x-4y=9. 3x − 4y = 9 3 x - 4 y = 9. Find the x-intercepts. Tap for more steps... To find the x-intercept (s), substitute in 0 0 for y y and solve for x x. 3 x − 4 ⋅ 0 = 9 3 x - 4 ⋅ 0 = 9. Solve the equation. Tap for more steps... Simplify 3 x − 4 ⋅ 0 3 x - 4 ⋅ 0.

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3 years ago

Solve the inequality<br> 7+7(x+4)>21

Kryger [21]

Answer:

x>-2

Step-by-step explanation:

Hi there!

We are given the inequality 7+7(x+4)>21, and we want to solve it

Solving an inequality is a lot like solving an equation; we want to isolate x on one side by itself.

So to start, we'll preform the distributive property on 7(x+4) (multiply x by 7, then multiply 4 by 7 and add those two together).

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Now combine like terms.

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Subtract 35 from both sides

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Divide both sides by 7

x>-2

Hope this helps!

8 0

3 years ago

Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -

GaryK [48]

Consider all parabolas:

$y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.$

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

$y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.$

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

$y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.$

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

$y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.$

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

$y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.$

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

$y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.$

When x=2, y=13, then the point (2,13) is vertex of this sixth parabola.

3 0

3 years ago