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2 years ago

What number is less than -2 4/5 and greater than -31/5

Mathematics

1 answer:

IgorC [24]2 years ago

8 0

Using a calculator, you should find that

-2 & 4/5 = -(2 + 4/5) = -(2+0.8) = -2.8

-31/5 = -6.2

So the unknown number is between -6.2 and -2.8

<h3>Possible whole number answers could be: -6, -5, -4, -3, or -2</h3>

If your teacher allows decimal answers, then there are infinitely many possible ways to answer.

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PLEASE HELPPPPP!!!!!!!!!!!!!!! SHOW WORK

larisa [96]

Answer:

17. A. 180 in²
18. B. 234 in²

Step-by-step explanation:

#17

<u>Total surface area is:</u>

A = 8(3*2 + 6*2) + 2(3*6) = 180 in²

Correct option is A

#18

<u>Total surface area is:</u>

A = 2(9*9) + 4(2*9) = 234 in²

Correct option is B

5 0

2 years ago

I NEED HELP LIKE RIGHT NOW!!

Bess [88]

Answer:

The answer in the attached figure

Step-by-step explanation:

Let

x------> the abscissa

y-----> the ordinate

we know that

$y=2x$ -----> given problem

The domain is $[-1, 0, 1]$

For $x=-1$

$y=2(-1)=-2$

The point is $(-1,-2)$

For $x=0$

$y=2(0)=0$

The point is $(0,0)$

For $x=1$

$y=2(1)=2$

The point is $(1,2)$

The answer in the attached figure

7 0

2 years ago

Read 2 more answers

Solve each equation y-6.72=253

diamong [38]

You just need to have the variable by itself, so move the -6.72 to the other side (that will make it +6.72)
So you should have this equation now: y = 253+6.72
Then just add them and you have this: y = 259.72
And that is your answer!

Hope I helped! <3

7 0

3 years ago

Read 2 more answers

find the values of the six trigonometric functions for angle theta in standard position if a point with the coordinates (1, -8)

frutty [35]

Answer:

cosФ = $\frac{1}{\sqrt{65}}$ , sinФ = $-\frac{8}{\sqrt{65}}$ , tanФ = -8, secФ = $\sqrt{65}$ , cscФ = $-\frac{\sqrt{65}}{8}$ , cotФ = $-\frac{1}{8}$

Step-by-step explanation:

If a point (x, y) lies on the terminal side of angle Ф in standard position, then the six trigonometry functions are:

cosФ = $\frac{x}{r}$
sinФ = $\frac{y}{r}$
tanФ = $\frac{y}{x}$
secФ = $\frac{r}{x}$
cscФ = $\frac{r}{y}$
cotФ = $\frac{x}{y}$

Where r = $\sqrt{x^{2}+y^{2} }$ (the length of the terminal side from the origin to point (x, y)

You should find the quadrant of (x, y) to adjust the sign of each function

∵ Point (1, -8) lies on the terminal side of angle Ф in standard position

∵ x is positive and y is negative

→ That means the point lies on the 4th quadrant

∴ Angle Ф is on the 4th quadrant

∵ In the 4th quadrant cosФ and secФ only have positive values

∴ sinФ, secФ, tanФ, and cotФ have negative values

→ let us find r

∵ r = $\sqrt{x^{2}+y^{2} }$

∵ x = 1 and y = -8

∴ r = $\sqrt{x} \sqrt{(1)^{2}+(-8)^{2}}=\sqrt{1+64}=\sqrt{65}$

→ Use the rules above to find the six trigonometric functions of Ф

∵ cosФ = $\frac{x}{r}$

∴ cosФ = $\frac{1}{\sqrt{65}}$

∵ sinФ = $\frac{y}{r}$

∴ sinФ = $-\frac{8}{\sqrt{65}}$

∵ tanФ = $\frac{y}{x}$

∴ tanФ = $-\frac{8}{1}$ = -8

∵ secФ = $\frac{r}{x}$

∴ secФ = $\frac{\sqrt{65}}{1}$ = $\sqrt{65}$

∵ cscФ = $\frac{r}{y}$

∴ cscФ = $-\frac{\sqrt{65}}{8}$

∵ cotФ = $\frac{x}{y}$

∴ cotФ = $-\frac{1}{8}$

8 0

2 years ago

a parabola has an x-intercept at 2, its axis of symmetry is the line x=4, and the y-coordinate of its vertex is 6. Determine the

nlexa [21]

Answer:

The standard equation of the parabola is:

$y=-\frac{3}{2}x^2+12x-18$

Step-by-step explanation:

An x intercept of 2 means that the point (2, 0) is in the graph of the parabola.

We can also write the general expression for the parabola in vertex form, since we can use the information on the coordinates of the vertex: (4, 6) - recall that the axis of symmetry of the parabola goes through the parabola's vertex, so the x-value of the vertex must be x=4.

$y-y_{vertex}=a\,(x-x_{vertex})^2\\y-6=a\,(x-4)^2$

Now we can find the value of the parameter "a" by using the extra information about the point (2, 0) at which the parabola intercepts the x-axis:

$y-6=a\,(x-4)^2\\0-6=a\,(2-4)^2\\-6=a\,4\\a=-\frac{6}{4} =-\frac{3}{2}$

Then the equation of the parabola becomes:

$y-6=-\frac{3}{2} \,(x-4)^2\\y-6=-\frac{3}{2} (x^2-8x+16)\\y-6=-\frac{3}{2}x^2+12x-24\\y=-\frac{3}{2}x^2+12x-18$

7 0

3 years ago