All categories

2 years ago

how do I put -1.2 on a number line?how do I put -3.25 on a number line?how do I put 0.5 on a number line?

Mathematics

1 answer:

ArbitrLikvidat [17]2 years ago

8 0

Answer:

-3.25,-1.2 then after 0 you put 0.5

Step-by-step explanation:

-3 -2.5 -1 -0.5 0 0.5 1 1.5 2 2.5

___l__l__l__l__l__l__l__l__l__l_____________

-3.25 -1.2 0.5

You might be interested in

9 hours _____ minutes

LiRa [457]

9 hours is 540 minutes

5 0

2 years ago

Read 2 more answers

Lena is asked to write an explicit formula for the graphed geometric sequence (3,6.25) (2,2.5) (1,1)

harkovskaia [24]

The correct answer for this question is this one: "2.5"
Lena is asked to write an explicit formula for the graphed geometric sequence. The value, written as a decimal, should Lena use as the common ratio that is 2.5There exists the same question from other source with the image is shown.
Hope this helps answer your question and have a nice day ahead.

3 0

3 years ago

Read 2 more answers

Find , , and if and terminates in quadrant .

ioda

sin2x =12/13

cos2x = 5/13

tan2x = 12/5

STEP - BY - STEP EXPLANATION

What to find?

• sin2x

• cos2x

• tan2x

Given:

tanx = 2/3 = opposite / adjacent

We need to first make a sketch of the given problem.

Let h be the hypotenuse.

We need to find sinx and cos x, but to find sinx and cosx, first determine the value of h.

Using the Pythagoras theorem;

hypotenuse² = opposite² + adjacent²

h² = 2² + 3²

h² = 4 + 9

h² =13

Take the square root of both-side of the equation.

h =√13

This implies that hypotenuse = √13

We can now proceed to find the values of ainx and cosx.

Using the trigonometric ratio;

$\sin x=\frac{opposite}{\text{hypotenuse}}=\frac{2}{\sqrt[]{13}}$ $\cos x=\frac{adjacent}{\text{hypotenuse}}=\frac{3}{\sqrt[]{13}}$

And we know that tanx =2/3

From the trigonometric identity;

sin 2x = 2sinxcosx

Substitute the value of sinx , cosx and then simplify.

$\sin 2x=2(\frac{2}{\sqrt[]{13}})(\frac{3}{\sqrt[]{13}})$ $=\frac{12}{13}$

Hence, sin2x = 12/13

cos2x = cos²x - sin²x

Substitute the value of cosx, sinx and simplify.

$\begin{gathered} \cos 2x=(\frac{3}{\sqrt[]{13}})^2-(\frac{2}{\sqrt[]{13}})^2 \\ \\ =\frac{9}{13}-\frac{4}{13} \\ =\frac{5}{13} \end{gathered}$

Hence, cos2x = 5/13

tan2x = 2tanx / 1- tan²x

$\tan 2x=\frac{2\tan x}{1-\tan ^2x}$ $=\frac{2(\frac{2}{3})}{1-(\frac{2}{3})^2}$ $=\frac{\frac{4}{3}}{1-\frac{4}{9}}$ $=\frac{\frac{4}{3}}{\frac{9-4}{9}}$ $=\frac{\frac{4}{3}}{\frac{5}{9}}$ $=\frac{4}{3}\times\frac{9}{5}=\frac{4}{1}\times\frac{3}{5}=\frac{12}{5}$