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3 years ago

Ahh I’m dumb, but I need help please!

Mathematics

2 answers:

natali 33 [55]3 years ago

6 0

Derek did finish first but 1/6=.16 so it woulda been 35.16.
I’m sure it was just a typo :)

andreev551 [17]3 years ago

3 0

Answer:

Derik finished first

Step-by-step explanation:

Derik = 35 1/6 = 35.10

other girl = 35.25

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What is 300,567,672,468.78 - 200.67

Marrrta [24]

300567672468.78-200.67=
300567672268.11005

4 0

3 years ago

Eight more than the product of a number and six equals three

mixas84 [53]

Answer:

8x + 6 = 3

Step-by-step explanation if solving for x

Simplifying

8x + 6 = 3

Reorder the terms:

6 + 8x = 3

Solving

6 + 8x = 3

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-6' to each side of the equation.

6 + -6 + 8x = 3 + -6

Combine like terms: 6 + -6 = 0

0 + 8x = 3 + -6

8x = 3 + -6

Combine like terms: 3 + -6 = -3

8x = -3

Divide each side by '8'.

x = -0.375

Simplifying

x = -0.375

3 0

4 years ago

Read 2 more answers

The height of a tree in feet over x years is modeled by the function f(x).

Evgesh-ka [11]

If e-the initial hight of the tree, and the function imply some brackets like

f(x)=30/(1+29e-0.5x) then

for every

$0.5x > 29e + 1$

the rate of growth is negative, the maximum growth per unit of time would be 30ft

$x \geqslant 1$

then

$29e < - 0.5$

so the tree couldn't be 2ft tall initially

if the brackets are something like

f(x)=30:(1+29e)-0.5x then the results may differ

8 0

3 years ago

This one is number 6

iren2701 [21]

I’m not entirely sure, but i do think it could be B.

5 0

2 years ago

Read 2 more answers

What is the 4th term of the expanded binomial (2x – 3y)^6

san4es73 [151]

Answer:

The 4th term of the expanded binomial is $-4320x^3y^3$

Step-by-step explanation:

Considering:

$(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k}y^k$

$(2x-3y)^6 = \sum_{k=0}^{6} \binom{6}{k} (2x)^{6-k}(-3y)^k$

Now, you gotta calculate for every value of $k$

$(2x-3y)^6 = \binom{6}{0} (2x)^{6-0}(-3y)^0 + \binom{6}{1} (2x)^{6-1}(-3y)^1 + \binom{6}{2} (2x)^{6-2}(-3y)^2 + \\$

$\binom{6}{3} (2x)^{6-3}(-3y)^3 + \binom{6}{4} (2x)^{6-4}(-3y)^4 + \binom{6}{5} (2x)^{6-5}(-3y)^5 + \binom{6}{6} (2x)^{6-6}(-3y)^6$

I will not write every product, but just solve following the steps:

For $k=0$

$\binom{6}{0} (2x)^{6-0}(-3y)^0$

$\frac{6!}{(6-0)!(0!)} (2x)^{6-0}(-3y)^0$

$\frac{6!}{6!} \left(2x\right)^{6-0}\cdot 1$

$1\cdot \:1\cdot \left(2x\right)^{6-0}$

$2^6x^6$

$64x^6$

$(2x-3y)^6=64x^6-576x^5y+2160x^4y^2-4320x^3y^3+4860x^2y^4-2916xy^5+729y^6$

8 0

3 years ago