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2 years ago

What does -16-3 equal

Mathematics

2 answers:

Airida [17]2 years ago

7 0

Answer:

Step-by-step explanation:

it equals -19

katrin2010 [14]2 years ago

3 0

Answer:

The value of the given expression $-16-3=-19$

Step-by-step explanation:

Given expression is $-16-3$

To find the value of the given expression:

$-16-3=-16-3$

Now taking the negative sign (-) outside the terms of the above equation we get,

$-16-3=-(16+3)$

Now apply the algebraic sum to the above expression we get,

$=-19$

Therefore $=-19$

The value of the given expression is $-16-3=-19$

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Subtract. Your answer should be a polynomial in standard form. (6y^2-9y+4) -(-7y^2+5y+1)=(6y 2 −9y+4)−(−7y 2 +5y+1)=left parenth

kifflom [539]

Solution:
(6y^2-9y+4)-(-7y^2+5y+1)
Distribute to get rid of ( )
6y^2-9y+4+7y^2-5y-1
Combine like terms
13y^2-14y+3

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3 years ago

Andrew owns stock in Hyren Airlines and as a result gets paid dividends worth $539.70 every year. If Andrew owns 105 shares of H

inn [45]

The correct answer for the question that is being presented above is this one: "b. $2.78." Andrew owns stock in Hyren Airlines and as a result gets paid dividends worth $539.70 every year. If Andrew owns 105 shares of Hyren Airlines, the dividend of each share yields <span>b. $2.78</span>

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3 years ago

Read 2 more answers

Pls help me!!

vlada-n [284]

Answer:

The graph of y = f(x), has vertex at (1, -2)

<u>The vertex of a function f(x - 3) is going to be:</u>

(1 - 3, -2) = (-2, -2)

<em>The graph of y = f(x) has the line x = 5 as an axis of symmetry. The graph also passes through the point (8,-7). Find another point that must lie on the graph of y = f(x).</em>

The axis of symmetry is at the same distance from the symmetric points.

x = 5 is a vertical line. The point (8, -7) is 3 units to the right. So the mirror point will be 3 units to the left and have same y-coordinate: x = 5 - 3 = 2

The point is (2, -7)

The graph in blue is the translation of the red to the left by 2 units.

<u>So the equation is:</u>

y = f(x + 2)

y = h(x) is graphed

h(7) = 5
h(h(7)) = h(5) = -1

The graph of the function y = u(x) given

This is a odd function.

The coordinates of u(x) and u(-x) add to zero because u(-x) = -u(x)

<u>Therefore:</u>

u(-2.72) + u(-0.81) + u(0.81) + u(2.72) =
[u(-2.72) + u(2.72)] + [u(-0.81) + u(0.81)] =
0 + 0 = 0

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3 years ago

13x+15=11x-25<br><br> 5t-5=5t+7<br><br> Please show steps for both <br> Thank you

Tanzania [10]

I hope this helps you

13x-11x=-25-15

2x=-40

x=-20

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2 years ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
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$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

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3 years ago