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matrenka [14]
2 years ago
7

How to subtract negative number from 1

Mathematics
2 answers:
bija089 [108]2 years ago
8 0

To subtract a negative number from one you will need to set it up like this

1- (-x) x equals the negative number you're talking about

so we use the method KCC (keep change change)

So we KEEP the positive one, CHANGE the subtraction sign in to a plus sign "+" then CHANGE the negative number to a positive

lets say the negative number is 2

1-(-2)

1+ (+2)

1+2 = 3

djyliett [7]2 years ago
3 0

when you subtract a negative number from any platitude number the equation changes to addition so you would add the value of the negative number to 1.

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2. If f(x) = 2X^2 + 3X Find f(-5)
IgorLugansk [536]

Answer:

35

Step-by-step explanation:

2(-5)^2+3(-5)

=2(25)+(-15)

=50-15

=35

PLS GIVE BRAINLIEST

4 0
2 years ago
How to solve-2(p+-8)+-2=6
natka813 [3]

Answer:

p=4

Step-by-step explanation:

first change all the addition signs in to minus if the numbers are negative.    ex. p+-8 = p-8.

So the equation equals to -2(p-8)-2=6.

Next +2 to both sides to cancel out the -2 and you get -2(p-8)=8

Divide both sides by -2 and you get (p-8)=-4. At this point you can remove the parenthesis. So the equation is p-8=-4

Finally +8 to both sides and you get p=4

3 0
3 years ago
Read 2 more answers
Help please!
bagirrra123 [75]

(d):  y = mx+n

m = -2/3 ⇒ y = (-2/3)x +n

A(-4,  6) ∈ d  ⇒ 6 = (-2/3)·(-4) +n ⇒ 6 = 8/3 +n ⇒

⇒ n = 6 - 8/3 ⇒ n = 10/3

Now, we have:

y = (-2/3)x +10/3


6 0
3 years ago
Evaluate 64 divided by [4 x (27 - 10)]
KatRina [158]
Using PEMDAS,
27-10=17
17*4=68
64/68=0.94 (rounded to the nearest hundredth) 
8 0
3 years ago
Read 2 more answers
We did not find results for: A measure of​ malnutrition, called the​ pelidisi, varies directly as the cube root of a​ person's w
asambeis [7]
\bf \qquad \qquad \textit{double proportional variation}\\\\
\begin{array}{llll}
\textit{\underline{y} varies directly with \underline{x}}\\
\textit{and inversely with \underline{z}}
\end{array}\implies y=\cfrac{kx}{z}\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad  variation
\end{array}\\\\
-------------------------------\\\\

\bf \begin{cases}
p=pedalisi\\
w=weight\\
s=\textit{sitting height}
\end{cases}\quad 
\begin{array}{llll}
%pelidisi, varies directly as the cube root of a​ person's weight in grams and inversely as the​ person's sitting height in centimeters.
\textit{pelidisi varies directly}\\
\textit{as cube root of weight}\\
\textit{and inversely to }\\
\textit{sitting height}
\end{array}\implies p=\cfrac{k\sqrt[3]{w}}{s}\\\\
-------------------------------

\bf \textit{we know that }
\begin{cases}
w=48,820\\
s=78.7\\
p=100
\end{cases}\implies 100=\cfrac{k\sqrt[3]{48820}}{78.7}
\\\\\\
100\cdot 78.7=k\sqrt[3]{48820}\implies \cfrac{7870}{\sqrt[3]{48820}}=k
\\\\\\
thus\qquad \boxed{p=\cfrac{\frac{7870}{\sqrt[3]{48820}}\sqrt[3]{w}}{s}}
\\\\\\
\textit{now, what is \underline{p} when }
\begin{cases}
w=54,688\\
s=72.6
\end{cases}?\implies p=\cfrac{\frac{7870}{\sqrt[3]{48820}}\sqrt[3]{54688}}{72.6}

now, if that value is less than 100, then the fellow is "undernourished", otherwise, is overfed.
3 0
3 years ago
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