All categories

3 years ago

Make a conjecture about the next item in the sequence 1, -4, 9, -16

2 answers:

4 0

To get -4 from 1, we must either subtract 5 from 1 or mult. 1 by -4. Unfortunately subtracting 5 from -4 does not result in 9, and mult -4 by -4 does not result in -16. So neither "works" in predicting the next term.

Can you think of any other experiments to do that would help you discover the pattern 1, -4, 9, -16?

HACTEHA [7]3 years ago

4 0

The absolute values of each term are a sequence of perfect squares and the sign alternates so the next term would be 5^2 = 25 answer.

You might be interested in

3. The Luxor hotel, in Las Vegas, is a square pyramid with lateral faces consisting of black, glassed mirrored panes. The buildi

krek1111 [17]

Given:
Square pyramid with lateral faces.
646 ft wide at the base.
350 ft high.

Because of the term lateral faces, we need to get the lateral area of the square pyramid.

Lateral Area = a √a² + 4 h² ; a = 646 ft ; h = 350 ft

L.A. = 646 ft √(646ft)² + 4 (350ft)²
L.A. = 646 ft √417,316 ft² + 4 (122,500 ft²)
L.A. = 646 ft √417,316 ft² + 490,000 ft²
L.A. = 646 ft √907,316 ft²
L.A. = 646 ft * 952.53 ft
L.A. = 615,334.38 ft²

3 0

3 years ago

Which line segment is a radius of circle A?

liubo4ka [24]

AB and AC ................................

7 0

3 years ago

Read 2 more answers

Question 4 of 10

Anvisha [2.4K]

Answer:

Step-by-step explanation:

I googled it

7 0

3 years ago

A. Evaluate ∫20 tan 2x sec^2 2x dx using the substitution u = tan 2x.

irakobra [83]

Answer:

The integral is equal to $5\sec^2(2x)+C$ for an arbitrary constant C.

Step-by-step explanation:

a) If $u=\tan(2x)$ then $du=2\sec^2(2x)dx$ so the integral becomes $\int 20\tan(2x)\sec^2(2x)dx=\int 10\tan(2x) (2\sec^2(2x))dx=\int 10udu=\frac{u^2}{2}+C=10(\int udu)=10(\frac{u^2}{2}+C)=5\tan^2(2x)+C$ . (the constant of integration is actually 5C, but this doesn't affect the result when taking derivatives, so we still denote it by C)

b) In this case $u=\sec(2x)$ hence $du=2\tan(2x)\sec(2x)dx$ . We rewrite the integral as $\int 20\tan(2x)\sec^2(2x)dx=\int 10\sec(2x) (2\tan(2x)\sec(2x))dx=\int 10udu=5\frac{u^2}{2}+C=5\sec^2(2x)+C$ .

c) We use the trigonometric identity $\tan(2x)^2+1=\sec(2x)^2$ is part b). The value of the integral is $5\sec^2(2x)+C=5(\tan^2(2x)+1)+C=5\tan^2(2x)+5+C=5\tan^2(2x)+C$ . which coincides with part a)

Note that we just replaced 5+C by C. This is because we are asked for an indefinite integral. Each value of C defines a unique antiderivative, but we are not interested in specific values of C as this integral is the family of all antiderivatives. Part a) and b) don't coincide for specific values of C (they would if we were working with a definite integral), but they do represent the same family of functions.

3 0

3 years ago

a company uses two vans to transport workers from a free parking lot to the workplace between 7:00 and 9:00 am . one van has 6 m

olganol [36]

Y - x = 6
y + 2x = 57

2y - 2x =12
y + 2x = 57

∴3y = 69
y = 23

23 - x = 6
-x = -17
x =17

large van = 23
small van = 17

4 0

3 years ago

Read 2 more answers