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mihalych1998 [28]

3 years ago

6

Ms. Flatau went to a garden shop and bought some potting soil for $17.50 and 4 shrubs. The total bill was $53.50. Write and solv

e an equation to find the price of each shrub.

2 answers:

yKpoI14uk [10]3 years ago

8 0

The equation would be (53.50-17.50)<span>÷4 because we need to subtract the potting soil, and since there are 4 shrubs, then you need to divide to get the total of one shrub. It could also be (s=shrubs) 4s + 17.50 = 53.50 and then solve for s. Now, let's solve it. 53.50-17.50=36. Now, divide 36 by 4, to get 9. So, each shrub cost $9. Hope this helped!</span>

Svetach [21]3 years ago

5 0

(53.50-17.50)<span>÷4 is what i think the anwser.

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a jacket originally sold for $45. this week it went on sale for 20% off. what is the discount and what is the sales price?

Strike441 [17]

Answer:

discount = 9

new price = 36

Step-by-step explanation:

The discount is the price times the discount percent

45 * 20%

Change to decimal form

45*.20

9

The new price is the original price minus the discount

45-9 = 36

4 0

3 years ago

Carmen rides her bicycle at a constant rate to the market. When she rides her bicycle back home along the same route, she bikes

prisoha [69]

We will use the equation:
r t = 3/4 r ( t+12 ) /:r
t = 3/4 t + 9
t - 3/4 t = 9
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4 years ago

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Aleks [24]

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5 0

3 years ago

Read 2 more answers

ANSWER ASAP PLEASE & SHOW UR WORK!!!

Lina20 [59]

Answer:

$x_1=\mathbf{i}\sqrt{6},\ x_2=-\mathbf{i}\sqrt{6},\ x_3=\sqrt{3},\ x_4=-\sqrt{3}$

Step-by-step explanation:

<u>Biquadratic Equations</u>

Solve:

$x^4 + 3x^2 - 18 = 0$

The biquadratic equations are equations of degree 4 without the terms of degree 1 and 3.

Solving such equations requires to express the equation as a second-degree equation with $x^2$ as the variable.

Rewriting the equation:

$(x^2)^2 + 3(x^2) - 18 = 0$

The quadratic equation can be factored as:

$(x^2+6)(x^2-3)=0$

It leads to two equations:

$x^2+6=0$

$x^2-3=0$

The first equation has imaginary roots. Solving for x:

$x^2=-6$

$x=\pm\sqrt{-6}$

$x_1=\mathbf{i}\sqrt{6}$

$x_2=-\mathbf{i}\sqrt{6}$

Where

$\mathbf{i}=\sqrt{-1}$

The second equation has two real roots:

$x^2-3=0$

$x^2=3$

$x=\pm\sqrt{3}$

$x_3=\sqrt{3}$

$x_4=-\sqrt{3}$

The roots are:

$\mathbf{x_1=\mathbf{i}\sqrt{6},\ x_2=-\mathbf{i}\sqrt{6},\ x_3=\sqrt{3},\ x_4=-\sqrt{3}}$

3 0

3 years ago

In triangle EFG , EF=FG . EF= 6X-10 , EG =2X+14, and FG =3X+11. find EF .

jarptica [38.1K]

Since EF=FG, you can set 6x - 10 = to 3x + 11

$6x -10 = 3x +11$

Then add like terms

$6x - 3x = 11+10$

so, $3x = 21$

Now you can divide by 3 to get $x = 7$ :)

then put it back into $6x - 10$

$6(7) - 10 = EF$

so, $EF = 32$

4 0

4 years ago