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blsea [12.9K]
2 years ago
15

The factorisation of y2-7y+12 is +100points and brainliest

Mathematics
2 answers:
damaskus [11]2 years ago
7 0
(y-4) (y-3)
you look at 12 and see what two factors of 12 add up to -7 and -4 times -3 is 12 and -4+-3 = -7
laiz [17]2 years ago
3 0
Ans= (y-3)(y-4).

Consider the form
x
2
+
b
x
+
c
x
2
+
b
x
+
c
. Find a pair of integers whose product is
c
c
and whose sum is
b
b
. In this case, whose product is
12
12
and whose sum is
−
7
-
7
.
−
4
,
−
3

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Rufina [12.5K]

The result of the exponential expression is 81

<h3>Indices expressions</h3>

Indices are written in term of exponents

According to the expoenent law of indices

a^4 = a * a * a * a

Given the expression

(-3)^4

This means the product of -3 in four places. Mathematically

(-3)^4 = -3 * -3 * -3 * -3
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(-3)^4 = 81

Hence the result of the exponential expression is 81

Learn more on exponents here: brainly.com/question/11975096

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Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st
Naily [24]

Answer:

\boxed{2^{\frac{802}{27}} \cdot 3^9}

Step-by-step explanation:

<u>I will try to give as many details as possible. </u>

First of all, I just would like to say:

\text{Use } \LaTeX !

Texting in Latex is much more clear and depending on the question, just writing down without it may be confusing or ambiguous. Be together with Latex! (*^U^)人(≧V≦*)/

$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

\left(\dfrac{1}{3^2}  \cdot \dfrac{1}{4^5}  \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3

Also,

\boxed{a^{0} = 1, a\neq 0 }

\left(\dfrac{1}{9}  \cdot \dfrac{1}{1024}  \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27

Note

\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq  0}

\left(\dfrac{1}{9216}   \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27

\left(\dfrac{1}{9216}   \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)

\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)

\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)

Note

\boxed{\dfrac{1}{\dfrac{1}{a} }  = a}

9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)

\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)

Once

9216=2^{10}\cdot 3^2 \implies  9216^3=2^{30}\cdot 3^6

\boxed{(a \cdot b)^n=a^n \cdot b^n}

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27} $

We have

\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)

Also, once

\boxed{\dfrac{c^a}{c^b}=c^{a-b}}

2^{30-\frac{8}{27}} \cdot 3^6\cdot 27

As

30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27}  =\dfrac{802}{27}

2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3

2^{\frac{802}{27}} \cdot 3^9

4 0
2 years ago
Graph step by step:<br> f(x)=-2x+1
Paul [167]

Step-by-step explanation:

Root:(1/2, 0)

Domain: x∈R

Range: y∈R

Vertical intercept: (0, 1)

4 0
2 years ago
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