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Serga [27]
2 years ago
8

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa chose all that apply

Mathematics
2 answers:
lora16 [44]2 years ago
6 0

Answer:

B,C

Step-by-step explanation:

aivan3 [116]2 years ago
4 0
B and c ................
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Identify the range of the function shown in the graph.
MAVERICK [17]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the plots and graphs.

Since, the x axis varies from -infinite to +infinite,

here, the y totally varies from -5 to +5,

hence, the option B.) is correct.

===>

- 5 \leqslant y \leqslant 5

3 0
2 years ago
What is the equation that represents the line that passes through the point (2,6) and has a slope of -5?
tiny-mole [99]

Answer:

y = - 5x + 16

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = - 5, thus

y = - 5x + c ← is the partial equation of the line

To find c substitute (2, 6) into the partial equation

6 = - 10 + c ⇒ c = 6 + 10 = 16

y = - 5x + 16 ← equation of line

3 0
3 years ago
A fishing supply store buys fishing rods for $11 and sells them at a markup of 220%. One day, they decide to put
allsm [11]

The answer is: $28.16

A markup of 220% would bring 11 to 35.20, and a mark down of 20% would bring 35.20 to 28.16.

Your'e Welcome  (ФДФ) .

4 0
2 years ago
Read 2 more answers
GIVING BRAINLIEST!!!!!
boyakko [2]

Answer:

A. Obtuse triangle

B,C,D. Acute

Step-by-step explanation:

3 0
2 years ago
Log base 2 (x-1) + log base 2 (x+9) = log base 2 (4x=3)
lisabon 2012 [21]
<span>using log property
</span>
log_b~(a)+log_b~(c)\Leftrightarrow log_b~(a*c)

and

log_b~(a)-log_b~(c)\Leftrightarrow log_b~(\frac{a}{c})

and

log_b~(a)=c\Leftrightarrow a=b^c


log_2~(x-1)+log_2~(x+9)=log_2~(4x+3)

log_2~[(x-1)*(x+9)]=log_2~(4x+3)

log_2~[(x-1)*(x+9)]-log_2~(4x+3)=0

log_2~\frac{(x-1)*(x+9)}{(4x+3)}=0

\frac{(x-1)*(x+9)}{(4x+3)}=2^0

now

(x-1)*(x+9)=(4x+3)

x^2+8x-9=4x+3

x^2+8x-9-4x-3=0

x^2+4x-12=0

\boxed{\boxed{X_1=-6~~and~~X_2=2}}

I hope you enjoy it...
5 0
3 years ago
Read 2 more answers
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