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il63 [147K]
3 years ago
5

A 5-lb box of soap powder costs $ 4.65. A 10-lb box of the same powder costs $ 8.90. Which is the better price?

Mathematics
1 answer:
siniylev [52]3 years ago
7 0

4.65 /5 = 0.93 per pound

8.90 /10 = 0.89 per pound

 the 10 pound box is the better price because it is cheaper per pound

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Slove algebraically<br><br>3x - 4y = -24<br>x +4y=8<br>​
Vanyuwa [196]

Solve algebrically 3x - 4y = -24 and x + 4y = 8 is x = -4 and y = 3

<u>Solution:</u>

We have been given two equations which are as follows:

3x - 4y = -24   ----- eqn 1

x + 4y = 8   -------- eqn 2              

We have been asked to solve the equations which means we have to find the value of ‘x’ and ‘y’.

We rearrange eqn 2 as follows:

x + 4y = 8

x = 8 - 4y  ------eqn 3

Now we substitute eqn 3 in eqn 1 as follows:

3(8 - 4y) -4y = -24

24 - 12y - 4y = -24

-16y = -48

y = 3

Substitute "y" value in eqn 3. Therefore the value of ‘x’ becomes:

x = 8 - 4(3)

x = 8 - 12  = -4

Hence on solving both the given equations we get the value of x and y as -4 and 3 respectively.

5 0
3 years ago
Two angles of a triangle are 30°and 80°. Find the third angle.
stiv31 [10]

Answer:

The third angle of the triangle is 70°.

Step-by-step explanation:

The two angles of a triangle (say Δ ABC) are given to be 30° and 80°.

We have to find the third angle.

Let us assume that ∠ A = 30° and ∠ B = 80°, then we have to find ∠ C.

Now, we know that, ∠ A + ∠ B + ∠ C = 180° {Property of a triangle}

⇒ 30° + 80° + ∠ C = 180°

⇒ ∠ C = 180° - 30° - 80° = 70°

Therefore, the third angle of the triangle is 70°. (Answer)

7 0
3 years ago
What is the horizontal displacement of the basic graph to produce a graph of
chubhunter [2.5K]

Since the equation says "cos(x-2)", we know that the horizontal displacement is 2 units.

3 0
3 years ago
Rearrange y=1/2x+2 to make x the subject.
yarga [219]

Subtract 2

y-2=1/2 x

Multiply by 2 to cancel out the denominator and to get 1 in front of x

2(y-2)=x

You can stop there or you can multiply out.

2y-4=x


7 0
3 years ago
Read 2 more answers
Can you help me solve an infinite geometric sequence
zaharov [31]

A geometric series like

\displaystyle \sum_{n=1}^\infty \dfrac{1}{\alpha^n}

Converges if and only if |\alpha|>1. If this is the case, the sum equals

\displaystyle \sum_{n=1}^\infty \dfrac{1}{\alpha^n} = \dfrac{1}{\alpha-1}

So, in your case, you have convergence if and only if

|2+a|>1 \iff 2+a>1 \lor 2+a-1 \lor a

And if this is the case, the sum equals

\displaystyle \sum_{n=1}^\infty \dfrac{1}{(2+a)^n} = \dfrac{1}{2+a-1} = \dfrac{1}{a+1}

8 0
3 years ago
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