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lbvjy [14]
3 years ago
6

Help!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
zhuklara [117]3 years ago
5 0

Answer:

see below

Step-by-step explanation:

9 ^-5 * 9^-3

When multiplying these we ADD the exponents:

9^-5 * 9^-3

= 9^(-5 + -3)

= 9 ^-8

= 1/9^8  Option A


2.    2^14 / 2^7   This is a division so we SUBTRACT the exponents:-

= 2^(14 - 7)  

= 2^7   (answer)

3.    If the triangle is a right angled one then it will obey Pythagoras Theorem so:-

13^ = x^2 + 5&2

x^2 = 13^2 - 5^2

x^2 = 144

x = 12    (answer)


4.   The last choice is a crucial step.

The total area of the triangles is the same in both large squares so  the area of the large  square e^2 = a^2 + b^2 ( the 2 squares in the left side large square).

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Solve<br><br> 4(3x - 2) = 52
guapka [62]

Answer:

x= 5

Step-by-step explanation:

first we open brackets

12x-8=52

then transpose

12x= 52+8

12x= 60

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hope it helps , pls mark me as brainliest

5 0
3 years ago
What is the vertex of the function?
egoroff_w [7]
<h2>Explanation:</h2><h2></h2>

Here you haven't provided any function, so I'll assume this one:

f(x)=3x^2-18x+33

This is a quadratic function whose graph represents a parabola. Any parabola has its vertex at the point:

V(x,y) \\ \\ \\ x=-\frac{b}{2a} \\ \\ y=f(-\frac{b}{2a}) \\ \\ \\ where: \\ \\ \\ f(x)=ax^2+bx+c \\ \\ \\ So: \\ \\ a=3 \\ \\ b=-18 \\ \\ c=33

Therefore:

x=-\frac{(-18)}{2(3)}=3 \\ \\ f(3)=3(3)^2-18(3)+33=27-54+33=6

Therefore, the vertex is:

\boxed{V(3,6)}

6 0
3 years ago
PLEASE HELPP MEE ASAPP!!
WARRIOR [948]

Answer:

Step-by-step explanation:

each zero of the function will have a factor of (x - x₀)

h(x) = a(x + 3)(x + 2)(x - 1)

h(x) = a(x + 3)(x² + x - 2)

h(x) = a(x³ + 4x² + x - 6)

or the third option works if a = 1

however this equation gives us the points (0, -6) and (-1. -4), so "a" must be -2

h(x) = -2x³ - 8x² - 2x + 12

to fit ALL of the given points as it fits the three zeros and also h(0) and h(-1) so I guess that is why the given group is a <u><em>partial</em></u> set of solution sets

7 0
3 years ago
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