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GarryVolchara [31]
3 years ago
8

Which property of equality is represented by the equation below?

Mathematics
2 answers:
oksian1 [2.3K]3 years ago
7 0
It is A. Identity property
valkas [14]3 years ago
3 0
Identity property I think
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Let R = {0, 1, 2, 3} be the range of h (x ) = x - 7. The domain of h inverse is
KonstantinChe [14]
Hello,


As h(x) is a bijection,
range(h(x))=dom(h^(-1)(x))={0,1,2,3}

Answer C
8 0
4 years ago
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What is the equation in point slope form of a line that passes through the point (−8, 2) and has a slope of 12 ? Drag a number,
NemiM [27]
Point-slope form:

y − b = m(x − a<span>), where m is the slope and (a, b) is a point on the graph.

Here that would look like:

y - 2 = 12(x - (-8))
y - 2 = 12(x + 8)

The answer:
</span>y - 2 = 12(x + 8)

Hope this helps.
5 0
4 years ago
Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k​
masha68 [24]

Answer:

Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

2

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

D=(1-k)^2-4(1)(k-3)D=(1−k)

2

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

2

−2k−4k+12

D=k^2-6k+13D=k

2

−6k+13

D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0
3 years ago
22 1/2% as a fraction
saveliy_v [14]
22 1/2 if this helps
7 0
3 years ago
Read 2 more answers
What fraction of the students in the class named baseball their favorite sport
mojhsa [17]

now, let's take a peek at the denominators, we have 3 and 8 and 12, we can get an LCD of 24 from that.

Let's multiply both sides by the LCD of 24, to do away with the denominators.

so, let's recall that a whole is "1", namely 500/500 = 1 = whole, or 5/5 = 1 = whole or 24/24 = 1 = whole.  So the whole class will yield a fraction of 1/1 or just 1.

\bf ~\hspace{7em}\stackrel{\textit{basketball}}{\cfrac{1}{3}}+\stackrel{\textit{soccer}}{\cfrac{1}{8}}+\stackrel{\textit{football}}{\cfrac{5}{12}}+\stackrel{\textit{baseball}}{x}~=~\stackrel{\textit{whole}}{1} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{24}}{24\left(\cfrac{1}{3}+\cfrac{1}{8}+\cfrac{5}{12}+x \right)=24(1)}\implies (8)1+(3)1+(2)5+(24)x=24 \\\\\\ 8+3+10+24x=24\implies 21+24x=24\implies 24x=3 \\\\\\ x=\cfrac{3}{24}\implies x=\cfrac{1}{8}

4 0
3 years ago
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