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3 years ago

A man spent exactly $2.50 on 3¢, 6¢, and 10¢ stamps. If he bought ten 3¢ stamps and

2 answers:

4 0

You have 2.50.
3 times 10 is 30, so now you have
2.20.
he bought twice as many six cent stamps as ten cent stamps.
multiples of six that are applicable are 30, 60, 90, 120.
six goes into 30 five times. Five can't be divided in half.
six goes into 60 ten times. ten divided in half is five. fifty plus sixty, is only 110.
six goes into ninety 15 times. fifteen can not be divided in half.
six goes into 120 twenty times. 20 divided in half is ten. ten times ten is 100.
100 + 120 = 220.
He bought 10 10cent stamps

stiks02 [169]3 years ago

3 0

1stamp i believe. Due to .03•10=.30 and 2.50-.30=2.20 so i divided By .6 and got 3.67 so just find where .10 fits. I know im no help

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A rectangle is a parallelogram with d. four congruent angles

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2 years ago

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1 year ago

Find the mode in 18 7 4 19 3 18 19 12

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3 years ago

A steel company sells 48 tons of steel each year.

ddd [48]

Answer:

16 tons

Step-by-step explanation:

To solve the first step is to change years to months because the time periods are in different units. There are 12 months in 1 year. This means that the company sells 48 tons over 12 months. 4 months is the equivalent of 1/3 of 12 months. This means that we can find the amount sold in 4 months by dividing 48 by 3. 48/3 is 16. Thus, the steel company sells 16 tons in 4 months.

Another way to find the answer is to find the unit rate. The unit rate is how many tons are sold in 1 month. Find this by dividing 48 by 12 because this represents the sales of 1 month. 48/12=4, this means that in 1 month, 4 tons of steel are sold. Next, multiply 4 times 4 to find the amount of steel sold in 4 months. 4 times 4 also gives you the same answer of 16.

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2 years ago

1+-w2+9w and I need help cuz I’m on 76 and I’m sooo close help

Gnesinka [82]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Simplify :- 1 + - w² + 9w.

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\large \sf1 + - w ^ { 2 } + 9 w$

Quadratic polynomial can be factored using the transformation $\sf \: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$ , where $\sf x_{1} and x_{2}$ are the solutions of the quadratic equation $\sf \: ax^{2}+bx+c=0$ .

$\large \sf-w^{2}+9w+1=0$

All equations of the form $\sf\:ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\sf\frac{-b±\sqrt{b^{2}-4ac}}{2a}$ . The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$\large \sf \: w=\frac{-9±\sqrt{9^{2}-4\left(-1\right)}}{2\left(-1\right)} \\$

Square 9.

$\large \sf \: w=\frac{-9±\sqrt{81-4\left(-1\right)}}{2\left(-1\right)} \\$

Multiply -4 times -1.

$\large \sf \: w=\frac{-9±\sqrt{81+4}}{2\left(-1\right)} \\$

Add 81 to 4.

$\large \sf \: w=\frac{-9±\sqrt{85}}{2\left(-1\right)} \\$

Multiply 2 times -1.

$\large \sf \: w=\frac{-9±\sqrt{85}}{-2} \\$

Now solve the equation $\sf\:w=\frac{-9±\sqrt{85}}{-2}$ when ± is plus. Add -9 to $\sf\sqrt{85}$ .

$\large \sf \: w=\frac{\sqrt{85}-9}{-2} \\$

Divide -9+ $\sf\sqrt{85}$ by -2.

$\large \boxed{ \sf \: w=\frac{9-\sqrt{85}}{2}} \\$

Now solve the equation $\sf\:w=\frac{-9±\sqrt{85}}{-2}$ when ± is minus. Subtract $\sf\sqrt{85}$ from -9.

$\large \sf \: w=\frac{-\sqrt{85}-9}{-2} \\$

Divide $\sf-9-\sqrt{85}$ by -2.

$\large \boxed{ \sf \: w=\frac{\sqrt{85}+9}{2}} \\$

Factor the original expression using $\sf\:ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$ . Substitute $\sf\frac{9-\sqrt{85}}{2}$ for $\sf\:x_{1}$ and $\sf\frac{9+\sqrt{85}}{2}$ for $\sf\:x_{2}$ .

$\large \boxed{ \boxed {\mathfrak{-w^{2}+9w+1=-\left(w-\frac{9-\sqrt{85}}{2}\right)\left(w-\frac{\sqrt{85}+9}{2}\right) }}}$

Well, in the picture you inserted it says that it's 8th grade mathematics. So, I'm not sure if you have learned simplification with the help of biquadratic formula. So, if you want the answer simplified only according to like terms then your answer will be ⇨

$\large \sf \: 1 + - w {}^{2} + 9w \\ =\large \boxed{\bf \: 1 - {w}^{2} + 9w}$

This cannot be further simplified as there are no more like terms (you can use the biquadratic formula if you've learned it.)

4 0

2 years ago