All categories

3 years ago

Gabby opened a savings account and deposited $800.00 as principal. The

Mathematics

1 answer:

Yuri [45]3 years ago

7 0

Answer:

Use the formula A = p x e^(rate x year), we have:

A = 800 x 2.71828^[(13/100) x 1] = 991.06 = ~911.1 dollar

Hope this helps!

You might be interested in

Is the number a multiple of8?Write yes or no.

Sergeeva-Olga [200]

Answer:

40 and 8 is the only multiple of 8

8x5=. 40

8x1 = 8

7 0

3 years ago

Read 2 more answers

Z varies directly with q. when z=2,q=1 . find a formula connecting z and q z=???

Lostsunrise [7]

Answer:

$z = \frac{q}{2}$

Step-by-step explanation:

$z \infty q$

$z = kq$

$z = 2 \: and \: q = 1$

$2 = k1$

$k = \frac{2}{1}$

$z = \frac{1}{2} q$

$z = \frac{q}{2}$

8 0

3 years ago

Read 2 more answers

Please help me fast!!!!!

AleksandrR [38]

Answer:

Step-by-step explanation:

If both factors are negative then the product is positive.

5 0

3 years ago

Read 2 more answers

Is 3/4 equivalent to 30/40

Liula [17]

No it is not the same

5 0

3 years ago

Read 2 more answers

Find the exact value of 6cos(105°)

Sveta_85 [38]

Answer:

$-\frac{3(\sqrt{6}-\sqrt{2})}{2}\text{ or } \frac{-3\sqrt{6}+3\sqrt{2}}{2}}\text{ or }\frac{3(\sqrt{2}-\sqrt{6})}{2}$

Step-by-step explanation:

There are multiple ways to achieve and even express the exact answer to this problem. Because the exact value of $6\cos(105^{\circ}})$ is a non-terminating (never-ending) decimal, it does not have a finite number of digits. Therefore, you cannot express it as an exact value as a decimal, as you'd either have to round or truncate.

Solution 1 (Cosine Addition Identity):

Nonetheless, to find the exact value we must use trigonometry identities.

Identity used:

$\cos(\alpha +\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta$

Notice that $45+60=105$ and therefore we can easily solve this problem if we know values of $\cos(45^{\circ})$ , $\cos(60^{\circ})$ , $\sin (45^{\circ})$ , and $\sin(60^{\circ})$ , which is plausible as they are all key angles on the unit circle.

Recall from either memory or the unit circle that:

$\cos(45^{\circ})=\sin(45^{\circ})=\frac{\sqrt{2}}{2}$

$\cos(60^{\circ})=\frac{1}{2}$
$\sin(60^{\circ})=\frac{\sqrt{3}}{2}$

Therefore, we have:

$\cos(105^{\circ})=\cos(45^{\circ}+60^{\circ}}),\\\cos(45^{\circ}+60^{\circ}})=\cos 45^{\circ}\cos 60^{\circ}-\sin 45^{\circ}\sin 60^{\circ},\\\cos(45^{\circ}+60^{\circ}})=\frac{\sqrt{2}}{2}\cdot \frac{1}{2}-\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2},\\\cos(105^{\circ})=\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4},\\\cos(105^{\circ})={\frac{-\sqrt{6}+\sqrt{2}}{4}}$

Since we want the value of $6\cos 105^{\circ}$ , simply multiply this by 6 to get your final answer:

$6\cdot {\frac{-\sqrt{6}+\sqrt{2}}{4}}=\frac{-3\sqrt{6}+3\sqrt{2}}{2}}=\boxed{\frac{3(\sqrt{2}-\sqrt{6})}{2}}$

Solution 2 (Combination of trig. identities):

Although less plausible, you may have the following memorized:

$\sin 15^{\circ}=\cos75^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4},\\\sin 75^{\circ}=\cos15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}$

If so, we can use the following trig. identity:

$\cos(\theta)=\sin(90^{\circ}-\theta)$ (the cosine of angle theta is equal to the sine of the supplement of angle theta - the converse is also true)

Therefore,

$\cos (105^{\circ})=\sin (90^{\circ}-105^{\circ})=\sin(-15^{\circ})$

Recall another trig. identity:

$\sin(-\theta)=-\sin (\theta)$ and therefore:

$\sin (-15^{\circ})=-\sin (15^{\circ})$

Multiply by 6 to get:

$6\cos (105^{\circ})=-6\sin (15^{\circ})=-6\cdot \frac{\sqrt{6}-\sqrt{2}}{4}=\boxed{-\frac{3(\sqrt{6}-\sqrt{2})}{2}}$ (alternative final answer).

4 0

3 years ago