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elena55 [62]
4 years ago
13

Helppppppp pleaseeeeeee

Mathematics
1 answer:
pshichka [43]4 years ago
4 0

Answer:

(2a, 2b)

Step-by-step explanation:

(0, 2b)

(2a, 0)

(2a, 2b)

I racked my brain for that.

You might be interested in
Graph the equation y=3x−6. Select two points to graph the line.
zhannawk [14.2K]

Answer:

Two points on the line are (0, -6) and (1, -3)

Step-by-step explanation:

I have a graphing calculator

5 0
3 years ago
Can someone please help i am very stuck :
Oliga [24]
The answer is
the first one is x= 18.98 
a^2+b^2=c^2
8 cm^2+17 cm^2= x^2 
64+289=x^2
√x^2= √353
x=18.89
<span><span>the second one is x= 18.7
a^2+b^2=c^2
7 m^2+ x^2= 20 m^2
49 m+x^2=400 m
-49            -49
  </span></span>√x^2=√352
<span><span>x= 18.7
Hope this helps!</span></span>
5 0
3 years ago
HELP PLS! THANK YOU SO MUCH! Consider the quadratic equation 3x^2-6=2x. (a) What is the value of the discriminant? (b) What does
ivanzaharov [21]

Answer:

see explanation

Step-by-step explanation:

Given a quadratic equation in standard form, ax² + bx + c = 0 ( a ≠ 0 )

Then the discriminant Δ = b² - 4ac informs us about the nature of the roots.

• If b² - 4ac > 0 then 2 real and distinct roots ( solutions )

• If b² - 4ac = 0 then 2 real and equal roots

• If b² - 4ac < 0 then roots are not real

Given

3x² - 6 = 2x ( subtract 2x from both sides )

3x² - 2x - 6 = 0 ← in standard form

with a = 3, b = - 2, c = - 6 , thus

b² - 4ac = (- 2)² - ( 4 × 3 × - 6) = 4 - (- 72) = 4 + 72 = 76

Since b² - 4ac > 0 then the solution is 2 real and distinct roots

8 0
3 years ago
If
sp2606 [1]

Answer:

51

Step-by-step explanation:

4x12-3x3+12

48-9+12=51

7 0
3 years ago
The value of a 1970 comic book has increased 12% per year. It originally sold for $0.35. Write an exponential equ Show your work
Alborosie

we can think of this as the same as a deposit in a bank, so let's use the compound interest formula, since the book is increasing in value 12% per annum.

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$0.35\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &t \end{cases} \\\\\\ A=0.35\left(1+\frac{0.12}{1}\right)^{1\cdot 5}\implies A=0.35(1.12)^t

6 0
3 years ago
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