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4 years ago

What is the 6th term of the geometric sequence 1024,528,256

Mathematics

1 answer:

zmey [24]4 years ago

3 0

167 i believe according to my calculations

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You have several purchases you need to make from the grocery store. from the grocery store you need to purchase 1/2 lb of cheese

MArishka [77]

The subtotal before tax and coupons is

... (2.99/lb × 1/2 lb) + (4.29/lb × 2 lb) + (2 loaves × 2.49/loaf) + (3 bags × 3.00/(2 bags)) + (2 salsa × 3.39/salsa) = 1.50 +8.58 +4.98 +4.50 +6.78

... = 26.34

This total is more than 25.00, but less than 30.00, so you get 3.00 off. The amount before tax is

... 26.34 - 3.00 = 23.34

4% tax is 0.04 × 23.34 = 0.93

so the total purchase comes to

... $23.34 + 0.93 = $24.27

3 0

3 years ago

Please helppppp??!!!???!

Stolb23 [73]

Answer:

Step-by-step explanation:

7 0

3 years ago

Write g(n)=2n(3-n)+5 in standard form

wolverine [178]

Standard form is an²+bn+c

expand
distribute
2n(3-n)+5=
2n(3)+2n(-n)+5=
6n-2n²+5=
-2n²+6n+5

g(n)=-2n²+6n+5 is standard form

6 0

4 years ago

9 - 3 ( 5 - 4 x ) =============

tangare [24]

Answer:26x will be your answer

8 0

3 years ago

Read 2 more answers

Hello! Verify the identity. Please show your work! Use trigonometric identities to verify each expression is equal.

RSB [31]

Answer:

See Below.

Step-by-step explanation:

We want to verify the identity:

$\displaystyle \csc^2 x -2\csc x \cot x +\cot ^2 x = \tan^2\left(\frac{x}{2}\right)$

Note that the left-hand side is a perfect square trinomial pattern. Namely:

$a^2-2ab+b^2=(a-b)^2$

If we let a = csc(x) and b = cot(x), we can factor it as such:

$\displaystyle (\csc x - \cot x)^2 = \tan^2\left(\frac{x}{2}\right)$

Let csc(x) = 1 / sin(x) and cot(x) = cos(x) / sin(x):

$\displaystyle \left(\frac{1}{\sin x}-\frac{\cos x }{\sin x}\right)^2=\tan^2\left(\frac{x}{2}\right)$

Combine fractions:

$\displaystyle \left(\frac{1-\cos x}{\sin x}\right)^2=\tan^2\left(\frac{x}{2}\right)$

Square (but do not simplify yet):

$\displaystyle \frac{(1-\cos x)^2}{\sin ^2x}=\tan^2\left(\frac{x}{2}\right)$

Now, we can make a substitution. Let u = x / 2. So, x = 2u. Substitute:

$\displaystyle \frac{(1-\cos 2u)^2}{\sin ^22u}=\tan^2u$

Recall that cos(2u) = 1 - sin²(u). Hence:

$\displaystyle \frac{(1-(1-2\sin^2u))^2}{\sin ^2 2u}=\tan^2u$

Simplify:

$\displaystyle \frac{4\sin^4 u}{\sin ^2 2u}=\tan^2 u$

Recall that sin(2u) = 2sin(u)cos(u). Hence:

$\displaystyle \frac{4\sin^4 u}{(2\sin u\cos u)^2}=\tan^2 u$

Square:

$\displaystyle \frac{4\sin^4 u}{4\sin^2 u\cos ^2u}=\tan^2 u$

Cancel:

$\displaystyle \frac{\sin ^2 u}{\cos ^2 u}=\tan ^2 u$

Since sin(u) / cos(u) = tan(u):

$\displaystyle \left(\frac{\sin u}{\cos u}\right)^2=\tan^2u=\tan^2u$

We can substitute u back for x / 2:

$\displaystyle \tan^2\left(\frac{x}{2}\right)= \tan^2\left(\frac{x}{2}\right)$

Hence proven.

3 0

3 years ago