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3 years ago

Evaluate −9 − (−4). 5 −5 13 −13

Mathematics

2 answers:

padilas [110]3 years ago

5 0

Answer:

$-5$

Step-by-step explanation:

We want to evaluate -9-(-4).

This is the same as $-9+4$ .

We can again rewrite this as $4-9$

Since we are subtracting a larger number from a smaller number, the result will be negative.

Therefore our expression evaluates to $-5$

The second option is correct.

elena-s [515]3 years ago

3 0

-5, because subtracting a negative is the same as adding a positive, which will bring -9 up to -5

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P + (-q) -3 when p= -6 and q=7

DochEvi [55]

Remember, -(-x)=x because the negatives cancel out

so
when p=-6 and q=7

p+(-q)-3
-6+(-7)-3
-6-7-3
-13-3
-16

5 0

4 years ago

Find the missing term and the missing coefficient 6a −( )5a =( ) a2 − 35a

andrew11 [14]

Solving for the missing term and the missing coefficient (6a − )5a = ( ) a^2 − 35a
Let the missing term be X
Let the missing coefficient be Y
Therefore, (6a – X)5a = Y(a^2) – 35a
6a x 5a – X.5a = Y.a^2 – 35a
30a^2 – X.5a = Y.a^2 – 35a
Equating co-efficients,
30a^2 = Y.a^2; X.5a = 35a
30 = Y; 5X = 35
Y = 30; X = 7
Therefore, (6a-7)5a = 30 a^2 – 35a <span>
</span>

3 0

3 years ago

The area of a rectangle is square units. Factor this expression. Given your answer in part A, describe what you can conclude abo

Brrunno [24]

Answer:

The factored area of the rectangle=2(2w-5)

Area of a rectangle=length×width

Step-by-step explanation:

Area of rectangle is 4w - 10 square units

We have to factor the expression

Area of rectangle = 4w - 10

common factor=2

Area of rectangle = 2(2w - 5)

Thus the given expression is factored

Area of a rectangle=length × width

Area of the rectangle=2×(2w-5)

Thus, it is safe to say

The length=2 or (2w-5)

The width=(2w-5) or 2

3 0

3 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago

PLEASE HELP MEEEE THANKS IF YIU DO

castortr0y [4]

B. Not moving, is the correct answer.

A is not correct because constant speed on a distance vs. time graph would vertically point to the right. C is not correct because acceleration on a distance vs. time graph would curve upwards. D is not correct because B applies to the graph.

Hope this helps :)

3 0

3 years ago