Combine like terms. 7a^2 - a^2 + 16 z^2 + z + 4z^3 + 4z^2 1/4(16 + 4p)

Mathematics

1 answer:

Ronch [10]3 years ago

5 0

1. -a^2+7x^2+16
2. 4z^3+5z^2+z
3. p+4

You might be interested in

The measure of one of the angles formed by two parallel lines and a transversal is 45°. Is it possible for the measure of any of

liubo4ka [24]

Answer: No

Step-by-step explanation:

Three of the other angles are 45°. The other four are 180° - 45°, which is not 55°. At the intersection of two lines, opposite angles are equal and adjacent angles are complementary, so two adjacent angles add to 180°. Adding a parallel line gives four more angles identical to the first four.

4 0

4 years ago

Write root of 2 correct to (a) 6 d.p

Solnce55 [7]

Answer:

1.414214

Step-by-step explanation:

Value is 1.41421<u>35</u>623730950488016887242097...

Since 7th digit is a 5, the 3 gets rounded to a 4 at the end.

6 0

3 years ago

Describe how you made a line plot in this lesson

serious [3.7K]

What lesson are we talking about I will be happy to help.

5 0

3 years ago

Read 2 more answers

If an unfair coin comes up heads 60% of the time and tails 40% of the time, what is the expectation if heads is valued at 1 a

monitta

I would think the value would be .2 cause if you had 100 tosses you would expect 60 value for heads and -40 value for tails. giving you 20/100

5 0

3 years ago

For what values of x does the curve y^2x^3-15x^2=4 have horizontal tangent lines

enyata [817]

We have the following curve:<span>

$y^2x^3-15x^2=4$

To find the tangent lines to the curve we need to use the concept of derivative, but we can’t solve this problem for $y$ , thus, let's apply implicit differentiation, so:

$\frac{d}{dx}(y^2x^3-15x^2=4) \\ \\
\frac{d}{dx}(y^2x^3)-\frac{d}{dx}(15x^2)=\frac{d}{dx}(4) \\ \\
(y^{2})'x^3+(x^3)'y^2-15(x^2)'=0 \\ \\ 2yy'x^3+3x^2y^2-30x=0 \\ \\
y'=\frac{dy}{dx}=\frac{30x-3x^2y^2}{2x^3y}=\frac{x(30-3xy^2)}{2x^3y} \\ \\
y'=\frac{30-3xy^2}{2x^2y}$

Therefore the horizontal lines occurs when $y'=0$ , then:

$\frac{30-3xy^2}{2x^2y}=0 \\ \\ That \ is, \
when: \\ \\ 30-3xy^2=0 \\ \\ \therefore xy^2=10 \rightarrow y^2=\frac{10}{x}$

If we substitute this in the original equation we have:

$\frac{10}{x}x^3-15x^2=4 \\ \\ \therefore
10x^2-15x^2=4 \\ \\ \therefore x^2=-\frac{4}{5}$

This is an absurd result because it is impossible for a squared number to get a negative number. So the conclusion is that there is no any value of $x$ in which the curve has horizontal tangent lines.</span>

8 0

3 years ago