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2 years ago

Choose the shape that has exactly three lines of symmetry.

Mathematics

2 answers:

saul85 [17]2 years ago

8 0

Answer:

an EQUILATERAL triangle!!

NikAS [45]2 years ago

3 0

Answer:

Equilateral Triangle

Step-by-step explanation:

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Can a person live a million days?

natka813 [3]

Answer:

I think so I'm not too sure

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3 years ago

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Chester Boles reviewed his credit card's monthly statement. It shows a $2,376.10 previous balance and this month's new purchases

ss7ja [257]

Initial balance, I = $2376.10 .

Total amount of purchase made, A = $( 875.22+65.75+45.22+21.23 ) = $1007.42 .

Total amount credit, c = $875.22 .

Fine, f = $45.30 .

Another purchase, $a=\dfrac{2376.10\times 2.5}{100}=\$ 59.4025$ .

So, balance left is :

B = I - A - f - a + c

B = 2376.10 - 1007.42 - 45.30 - 59.4025 + 875.22

B = $2139.1975

Hence, this is the required solution.

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2 years ago

after allowing 20 percent discount on the marked price of a radio 15 percent vat is levied on it , if its price become rs 22080

Kamila [148]

Answer:

Step-by-step explanation:

Hello, let's say that the price was P, a real number.

After 20% discount it become P - 20% * P = P* (1-20%) = P * (1 - 0.2)

= P * 0.8

And then we take 15% for the VAT, the new price become P * 0.8 * ( 1 + 15%)

= P * 0.8 * 1.15

And this is equal to 22080, so

P * 0.8 * 1.15 = 22080

and the amount of the VAT is P *0.8 * 0.15

$=\dfrac{22080}{1.15}*0.15=2880$

Hope this helps.

Thank you.

5 0

3 years ago

Whats is the distance between domican republic adn new york

Andrew [12]

1,534 miles. Good luck!

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3 years ago

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Write the equation of the quadratic function whose graph passes through <img src="https://tex.z-dn.net/?f=%28-3%2C2%29" id="TexF

blagie [28]

Answer:

$f(x)=x^2+3x+2$

Step-by-step explanation:

We want to write the equation of a quadratic whose graph passes through (-3, 2), (-1, 0), and (1, 6).

Remember that the standard quadratic function is given by:

$f(x)=ax^2+bx+c$

Since it passes through the point (-3, 2). This means that when $x=-3$ , $f(x)=f(-3)=2$ . Hence:

$f(-3)=2=a(-3)^2+b(-3)+c$

Simplify:

$2=9a-3b+c$

Perform the same computations for the coordinates (-1, 0) and (1, 6). Therefore:

$0=a(-1)^2+b(-1)+c \\ \\0=a-b+c$

And for (1, 6):

$6=a(1)^2+b(1)+c\\\\ 6=a+b+c$

So, we have a triple system of equations:

$\left\{ \begin{array}{ll} 2=9a-3b+c &\\ 0=a-b+c \\6=a+b+c \end{array} \right.$

We can solve this using elimination.

Notice that the b term in Equation 2 and 3 are opposites. Hence, let's add them together. This yields:

$(0+6)=(a+a)+(-b+b)+(c+c)$

Compute:

$6=2a+2c$

Let's divide both sides by 2:

$3=a+c$

Now, let's eliminate b again but we will use Equation 1 and 2.

Notice that if we multiply Equation 2 by -3, then the b terms will be opposites. So:

$-3(0)=-3(a-b+c)$

Multiply:

$0=-3a+3b-3c$

Add this to Equation 1:

$(0+2)=(9a-3a)+(-3b+3b)+(c-3c)$

Compute:

$2=6a-2c$

Again, we can divide both sides by 2:

$1=3a-c$

So, we know have two equations with only two variables:

$3=a+c\text{ and } 1=3a-c$

We can solve for a using elimination since the c term are opposites of each other. Add the two equations together:

$(3+1)=(a+3a)+(c-c)$

Compute:

$4=4a$

Solve for a:

$a=1$

So, the value of a is 1.

Using either of the two equations, we can now find c. Let's use the first one. Hence:

$3=a+c$

Substitute 1 for a and solve for c:

$\begin{aligned} c+(1)&=3 \\c&=2 \end{aligned}$

So, the value of c is 2.

Finally, using any of the three original equations, solve for b:

We can use Equation 3. Hence:

$6=a+b+c$

Substitute in known values and solve for b:

$6=(1)+b+(2)\\\\6=3+b\\\\b=3$

Therefore, a=1, b=3, and c=2.

Hence, our quadratic function is:

$f(x)=x^2+3x+2$

5 0

2 years ago