All categories

4 years ago

Write an equation please help

Mathematics

1 answer:

maks197457 [2]4 years ago

8 0

If she picks up 1/2 of a box for 6 days, then you would multiply 1/2 by 6. And add the 1 box she started out with. (1/2 × 6) +1 = 4

You might be interested in

Erin opened a savings account that earns 3% simple interest. If she deposts $2,750 and doesnt make any additional deposits or wi

irga5000 [103]

Answer:

3245$ (im assuming the interest is yearly)

Step-by-step explanation:

using the is/of method

3% x

___ ___ cross mult and divide. = 82.5 interest x 6 years of it

100 2750

2750 + 495

8 0

3 years ago

What is the 4th term of the expanded binomial (2x – 3y)^6

san4es73 [151]

Answer:

The 4th term of the expanded binomial is $-4320x^3y^3$

Step-by-step explanation:

Considering:

$(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k}y^k$

$(2x-3y)^6 = \sum_{k=0}^{6} \binom{6}{k} (2x)^{6-k}(-3y)^k$

Now, you gotta calculate for every value of $k$

$(2x-3y)^6 = \binom{6}{0} (2x)^{6-0}(-3y)^0 + \binom{6}{1} (2x)^{6-1}(-3y)^1 + \binom{6}{2} (2x)^{6-2}(-3y)^2 + \\$

$\binom{6}{3} (2x)^{6-3}(-3y)^3 + \binom{6}{4} (2x)^{6-4}(-3y)^4 + \binom{6}{5} (2x)^{6-5}(-3y)^5 + \binom{6}{6} (2x)^{6-6}(-3y)^6$

I will not write every product, but just solve following the steps:

For $k=0$

$\binom{6}{0} (2x)^{6-0}(-3y)^0$

$\frac{6!}{(6-0)!(0!)} (2x)^{6-0}(-3y)^0$

$\frac{6!}{6!} \left(2x\right)^{6-0}\cdot 1$

$1\cdot \:1\cdot \left(2x\right)^{6-0}$

$2^6x^6$

$64x^6$

$(2x-3y)^6=64x^6-576x^5y+2160x^4y^2-4320x^3y^3+4860x^2y^4-2916xy^5+729y^6$

8 0

3 years ago

Match each equation on the left with the number and type of its solutions on the right.

klemol [59]

Answer:

Step-by-step explanation:

1). Given equation is,

2x² - 3x = 6

2x² - 3x - 6 = 0

To find the solutions of the equation we will use quadratic formula,

x = $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute the values of a, b and c in the formula,

a = 2, b = -3 and c = -6

x = $\frac{3\pm\sqrt{(-3)^2-4(2)(-6)}}{2(2)}$

x = $\frac{3\pm\sqrt{9+48}}{4}$

x = $\frac{3\pm\sqrt{57}}{4}$

x = $\frac{3+\sqrt{57}}{4},\frac{3-\sqrt{57}}{4}$

Therefore, there are two real solutions.

2). Given equation is,

x² + 1 = 2x

x² - 2x + 1 = 0

(x - 1)² = 0

x = 1

Therefore, there is one real solution of the equation.

3). 2x² + 3x + 2 = 0

By applying quadratic formula,

x = $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

x = $\frac{-3\pm\sqrt{3^2-4(2)(2)}}{2(2)}$

x = $\frac{-3\pm\sqrt{9-16}}{4}$

x = $\frac{-3\pm i\sqrt{7}}{4}$

x = $\frac{-3+ i\sqrt{7}}{4},\frac{-3- i\sqrt{7}}{4}$

Therefore, there are two complex (non real) solutions.

3 0

3 years ago

NEED THE ANSWER FAST PLEASE!!!!!

dimaraw [331]

Answer:

26% because you multiply 26 by 100 and get 2,600 so thats 26% Hope this helps :)

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

The midpoint between y and 33 is -7. Find y

Basile [38]

Answer:

y = -47

Step-by-step explanation:

To find the midpoint of two points, add them together and divide by 2

(y+33)/2 = -7

To find y, multiply each side by 2

(y+33)/2 *2 = -7*2

y+33 = -14

Subtract 33 from each side

y+33-33 = -14-33

y = -47

7 0

3 years ago