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3 years ago

I'm getting so frustrated right now! I'm srry.

Mathematics

2 answers:

MissTica3 years ago

6 0

Y is plus 4 by each x number

Alchen [17]3 years ago

6 0

Y+4=x

Hope this helps ya

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Solve for x. MUST SHOW ALL WORK TO RECEIVE FULL CREDIT! (See picture for equation)

ioda

$\frac{x}{2.4} = \frac{3}{8}$

For proportions and to find x we have to cross multiply
8 × (x) = 3 × 2.4

8x = 7.2
Divide by 8 to isolate x

$\frac{8x}{8} = \frac{7.2}{8}$
8 and 8 cancels out

x = 0.9

5 0

3 years ago

Read 2 more answers

Need help with this problem

insens350 [35]

That equation just equals ln(e^(-4/3)) which can be simplified to -4/3 because ln and e cancel each other out only leaving the power

4 0

3 years ago

Keli has 26 roses and 73 carnations. She will put an assortment of 6 flowers in each vase. How many vases will she need for her

Irina18 [472]

Add 26 and 73 = 99 then divide 99 by 6 = 16.5 which is 17 vases

8 0

3 years ago

Figure 14-3

enot [183]

The answer is D. SIS idk how to do this
Ur welcome

5 0

3 years ago

Find the exact value of each trigonometric function for the given angle θ.

Kay [80]

Answer:

$\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=\dfrac{2}{\sqrt{3}}.$

Step-by-step explanation:

The given angle is 240 degrees.

We need to find the exact value of each trigonometric function for the given angle θ.

Since $\theta=240$ , it means θ lies in 3rd quadrant. In 3d quadrant only tan and cot are positive.

$\sin (240^\circ)=\sin (180^\circ+60^\circ)=-\sin (60^\circ)=-\dfrac{\sqrt{3}}{2}$

$\cos (240^\circ)=\cos (180^\circ+60^\circ)=-\cos (60^\circ)=-\dfrac{1}{2}$

$\tan (240^\circ)=\tan (180^\circ+60^\circ)=\tan (60^\circ)=\sqrt{3}$

$\cot (240^\circ)=\cot (180^\circ+60^\circ)=\cot (60^\circ)=\dfrac{1}{\sqrt{3}}$

$\sec (240^\circ)=\sec (180^\circ+60^\circ)=-\sec (60^\circ)=-2$

$\csc (240^\circ)=\csc (180^\circ+60^\circ)=-\csc (60^\circ)=-\dfrac{2}{\sqrt{3}}$

Therefore, $\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=-\dfrac{2}{\sqrt{3}}.$

8 0

3 years ago