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2 years ago

In June, you start a holiday savings account

Mathematics

2 answers:

Leona [35]2 years ago

6 0

Answer:

You will have $54 by December and here's why:

Step-by-step explanation:

You deposite $4 every month until December, that's 6 months.

so you do $4 × 6 months = 24

So you add 24 and 30

$24 + $30 = $54

jarptica [38.1K]2 years ago

3 0

Answer:

$54

Step-by-step explanation:

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If f(x)=3x-2,find f(8)-f(-5)

Alexandra [31]

To find both f(8) and f(-5), we will need to plug in their values for x and solve for each. Then, we can subtract them to find the final answer.

Finding f(8):

f(8) = 3(8) - 2

f(8) = 24 - 2

f(8) = 22

Finding f(-5):

f(-5) = 3(-5) - 2

f(-5) = -15 - 2

f(-5) = -17

Subtracting f(8) and f(-5):

f(8) - f(-5)

22 - - 17

22 + 17

Hope this helps!! :)

6 0

3 years ago

Does anyone know how to solve this?

KATRIN_1 [288]

Answer:

<30,-20>

Step-by-step explanation:

4u + 2v

4< 6,-4> + 2< 3,-2>

<24,-16> + < 6,-4>

<30,-20>

4 0

2 years ago

X + y = 152,<br><br> 8.5x + 12y = 1,656<br><br> How many hats were sold?

Elodia [21]

Answer:

x = 48 and y = 104

Step-by-step explanation:

Given equations are:

$x+y = 152\\8.5x+12y=1656$

From equation 1:

$x = 152-y$

Putting the value of y in equation 2

$8.5(152-y)+12y = 1656\\1292-8.5y+12y = 1656\\3.5y+1292 = 1656\\3.5y = 1656-1292\\3.5y = 364\\\frac{3.5y}{3.5} = \frac{364}{3.5}\\y = 104$

Now we have to put the value of y in one of the equation to find the value of x

Putting y = 104 in the first equation

$x+y = 152\\x+ 104 = 152\\x = 152-104\\x = 48$

Hence,

The solution of the system of equations is x = 48 and y = 104

The value of variable which was assumed for number of hats, is the total number of hats.

6 0

2 years ago

Isabella worked 50 hours over the past two weeks, and she gets paid by the hour. During the first week of the two-week span, she

4vir4ik [10]

Answer:

$190.00

Step-by-step explanation:

285÷30=9.5

9.5×20=190

Hope this helps! :)

8 0

3 years ago

What is the following product? (2 square root 7+3 square root 6)(5 square root 2+4 square root 3)

Xelga [282]

$(2\sqrt7+3\sqrt6)(5\sqrt2+4\sqrt3)\qquad\text{use distributive property}\\\\=(2\sqrt7)(5\sqrt2)+(2\sqrt7)(4\sqrt3)+(3\sqrt6)(5\sqrt2)+(3\sqrt6)(4\sqrt3)\\\\=10\sqrt{14}+8\sqrt{21}+15\sqrt{12}+12\sqrt{18}\\\\=10\sqrt{14}+8\sqrt{21}+15\sqrt{4\cdot3}+12\sqrt{9\cdot2}\\\\=10\sqrt{14}+8\sqrt{21}+15\sqrt4\cdot\sqrt3+12\sqrt9\cdot\sqrt2\\\\=10\sqrt{14}+8\sqrt{21}+(15)(2)\sqrt3+(12)(3)\sqrt2\\\\=\boxed{10\sqrt{14}+8\sqrt{21}+30\sqrt3+36\sqrt2}$

5 0

3 years ago

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