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2 years ago

Anna put $2 500 in a bank and left it for two years. At the end of the 2 years, she realized that she had $2.700 in her

Mathematics

1 answer:

user100 [1]2 years ago

4 0

Answer:

$ 100 a year

Step-by-step explanation:

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Patrick paid $18.70 for a pack of 10 notebooks. Individually, each notebook in the pack costs $ . PLS HELPPPPPP now

Cloud [144]

Answer:

$1.87

Step-by-step explanation:

Since the total cost was $18.70 and there are ten in the pack, to find the price of one notebook you just need to divide the total cost by the amount of notebooks there are. So 18.70 divided by 10 is $1.87 for each notebook.

5 0

2 years ago

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N 1968, it cost $54,000 for a 30-second commercial during the Super Bowl. In 1973, the cost for a 30-second commercial was $88,0

Yuri [45]

Answer:

In 5 years, the price increased about 30%.

I am not good at perentages, so correct me if I'm wrong. Also this is an estimation, not the exact percentage.

6 0

2 years ago

What is the value of the equation 9/2÷(-3)0

Finger [1]

The answer just keeps coming back to zero dude it's simple you can't really calculate this either way the value of it would just be zero you can't simplify it or anything

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2 years ago

PLEASE ANSWER!!!!!! REALLY IMPORTANT

faust18 [17]

Answer-

The height of the prism is 6 units

Solution-

As the base of the prism is a hexagon consisting of 2 congruent isosceles trapezoids.

So,

$V_{Prism}=Area_{Base}\times Height$

And,

$Area_{Base}=2\times \text{Area of the trapezoid}$

Also,

$\text{Area of the trapezoid}=\dfrac{1}{2}\times \text{Height}\times (\text{Sum of two parallel lines})$

$=\dfrac{1}{2}\times 3\times (5+8)\\\\=\dfrac{39}{2}$

Putting all the values,

$V_{Prism}=2\times \dfrac{39}{2}\times Height=39\times Height$

As the volume is given, so

$\Rightarrow 39\times Height=234$

$\Rightarrow Height=\dfrac{234}{39}=6$

8 0

2 years ago

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What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?

Ipatiy [6.2K]

Answer:

$x = \frac{ 5 \ + \ \sqrt{31}}{2} \ , \ x = \frac{ 5 \ - \ \sqrt{31}}{2}$

Step-by-step explanation:

$2x^2 - 10x - 3 = 0 \\\\a = 2 \ , b = - 10 \ , \ c = - 3 \\\\x = \frac{-b^2\ \pm \ \sqrt{b^2 - 4ac}}{2a}\\\\x = \frac{10 \ \pm \sqrt{(-10)^2 - ( 4 \times 2 \times -3)} }{2 \times 2}\\\\x = \frac{10 \ \pm \sqrt{(100 - ( -24 )} }{4}\\\\x = \frac{10 \ \pm \sqrt{(100 + 24 } }{4}\\\\x = \frac{ 10 \ \pm \sqrt{124}}{4}\\\\x = \frac{ 10 \ \pm \sqrt{4 \times 31}}{4}\\\\x = \frac{ 10 \ \pm \sqrt{2^2 \times 31}}{4}\\\\x = \frac{ 10 \ \pm2 \sqrt{31}}{4}\\\\x = \frac{ 5 \ \pm\sqrt{31}}{2}\\\\$

$x = \frac{ 5 \ + \ \sqrt{31}}{2} \ , \ x = \frac{ 5 \ - \ \sqrt{31}}{2}$

3 0

2 years ago