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Morgarella [4.7K]
3 years ago
9

Can you help me on his question

Mathematics
2 answers:
liraira [26]3 years ago
7 0
C 2 (-20)

He withdrew (took out) money twice.
That's -20 two times.
cestrela7 [59]3 years ago
3 0
How do you subtract an inter from another integer without using a number line or counters
You might be interested in
The finance department at a regional toy company has been tracking the income and costs of a new line of dolls. They have determ
romanna [79]

Answer:

q#1 Option B.2 possible solution is correct option

Q#2 option c. 1 viable solution is correct option.

Step-by-step explanation:

Q#1

y=4+3x+45

as this is a quadratic solution

and we know that when we solve a quadratic equation then  it gives two possible solutions

hence option b is the correct option

Q#2

option c is correct option when we solve an quadratic equation it gives two solution one is positive and other is negative as we know that income cannot be negative

hence only one viable solution exists when we solve this

y=4+3x+45 quadratic equation

Read more on Brainly.com - brainly.com/question/9837160#readmore

Step-by-step explanation:

3 0
3 years ago
How do I write a recursive formula for this sequence? 2, 6, 18, 54, 162, ... I already know the common difference is x3.
dmitriy555 [2]
For the sequence 2, 6, 18, 54, ..., the explicit formula is: an = a1 ! rn"1 = 2 ! 3n"1 , and the recursive formula is: a1 = 2, an+1 = an ! 3 . In each case, successively replacing n by 1, 2, 3, ... will yield the terms of the sequence. See the examples below.
4 0
3 years ago
I have the question in a screengrab. Thank you, whoever helps me.
nata0808 [166]

To simplify

\sqrt[4]{\dfrac{24x^6y}{128x^4y^5}}

we need to use the fact that

\sqrt[4]{x^4}=|x|

Why the absolute value? It's because (-x)^4=(-1)^4x^4=x^4.

We start by rewriting as

\sqrt[4]{\dfrac{2^23x^6y}{2^6x^4y^5}}

\sqrt[4]{\dfrac{2^23x^4x^2y}{2^42^2x^4y^4y}}

Since x\neq0, we have \dfrac xx=1, and the above reduces to

\sqrt[4]{\dfrac{3x^2y}{2^4y^4y}}

Then we pull out any 4th powers under the radical, and simplify everything we can:

\dfrac1{\sqrt[4]{2^4y^4}}\sqrt[4]{\dfrac{3x^2y}{y}}

\dfrac1{|2y|}\sqrt[4]{3x^2}

where y>0 allows us to write \dfrac yy=1, and this also means that |y|=y. So we end up with

\dfrac{\sqrt[4]{3x^2}}{2y}

making the last option the correct answer.

7 0
3 years ago
Where is the hole for the following function located?
Sholpan [36]
The hole of a function is described as the x-y coordinate wherein the denominator and numerator equates to zero. For the function f(x)= x+3/(x+4)(x+3), the common polynomial given is x-3 which can be equated to zero. Hence, 0 divided 0 means a hole in the function. If x+3 = 0, the hole of the function is at x = -3. ^w^
5 0
2 years ago
Please need help with this question m
katovenus [111]

Answer:

D.

Step-by-step explanation:

3(5)+2 > 14= 17>14 which is true

10-5 less than or equal to 7

true, 5 is less than 7

7 0
3 years ago
Read 2 more answers
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