Ask question

Ask question

All categories

Irina-Kira [14]

3 years ago

13

The grocery store sells kumquats for $5.00 a pound and Asian pears for $3.25 a pound. Write an equation in standard form for the

weights of kumquats k and Asia pears p that a customer could buy with $18

2 answers:

Vsevolod [243]3 years ago

8 0

Answer:

Equation will be 5k + 3.25p = $18

Step-by-step explanation:

Price of kumquats per pound = $5.00

Price of Asian pears per pound = $3.25

Let the weight of kumquats be represented by = k

Let the weight of Asian pears be represented by = p

As asked, how much kumquats and pears can be bought in $18

The equation will be 5k + 3.25p = $18

aniked [119]3 years ago

3 0

Okay so your equation is going to be : 18= 5k + 3.25p

You might be interested in

The 3rd term of a geometric sequence is -1/3 and the 8th term is 81. what is the 11th term?

shepuryov [24]

Answer: $t_{11}=-2187$

Step-by-step explanation:

$t_{3}=-1/3$

That is :

$ar^{2} = -1/3$ ................. equation 1

Also

$t_{8}=81$

that is

$ar^{7}=81$ ................. equation 2

divide equation 2 by equation 1 , that is

$\frac{ar^{7}}{ar^{2}} = 81 / -1/3$

$r^{5}=$ $81$ x $-3/1$

$r^{5}= -243$

find the fifth root of both sides

$r = -3$

substitute $r = -3$ , into equation 1 to find a , that is

$a(-3^{2})= -1/3$

$9a = -1/3$

multiply through by 3

$27a = -1$

$a =- 1/27$

To find the 11th term , the formula for the 11th term is given as :

$t_{11}=ar^{10}$

$t_{11}= -1/27(-3^{10})$

$t_{11}=-2187$

6 0

4 years ago

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

Isabel is shopping for her new car. She identified six models she likes at a local dealership. The table lists the down payment

tamaranim1 [39]

The amount in Isabel's account after 3 years can be calculated using the

future value of an annuity formula.

The cars within Isabel's budget are;

Choice 2

Choice 3

Choice 4

Reasons:

The amount she saves every month, PMT = $200

The compound interest on the account, r = 1.7%

The value of the account after t = 3 years is given as follows;

$\displaystyle FV = \mathbf{ PMT \cdot \frac{\left( \left( 1 + i \right)^n - 1 \right)}{i}}$

Where;

$\displaystyle i = \frac{r}{12}$

n = 12 × t

Which gives;

$\displaystyle FV = 200 \times \frac{\left( \left( 1 + \frac{1.7}{12} \right)^{12 \times 3} - 1 \right)}{\frac{1.7}{12} } \approx \mathbf{7,381.4}$

Therefore, the cars within her budget for which she can make a down payment are cars that require a down payment of less than $7,381.4, which are;

Choice 2, that requires a down payment of $6,200

Choice 3, that requires a down payment of $5,100

Choice 4, that require a down payment of $7,250

<em>Question parameters obtained from a similar question posted online are;</em>

<em>The amount Isabel saves each month = $200</em>

<em>The compound interest rate on the account = 1.7% compounded monthly</em>

Learn more about the future value of an annuity here:

brainly.com/question/4440017

7 0

3 years ago

Please help me with some math im stuck

swat32

y = - 40

- 40 since theres no letters it gonna be -40.

3 0

3 years ago

I only need help on #19. plz show work thanks.

Likurg_2 [28]

I'm not sure how to show work on this problem but hopefully this helps.

4 0

3 years ago