All categories

3 years ago

I need help with this ASAP! Its about budgeting.

Mathematics

1 answer:

Umnica [9.8K]3 years ago

3 0

Answer:

Step-by-step explanation:

I gave you all of the information to fill out the first page

i also answered some of the questions on the 2nd page. They are written sideways on the page. The other questions I didn’t answer yneed calculations after you add all of the expenses together.

put the amounts from each category I listed in the image and put them into the respective category and get the total for each category. Use those totals to fill out page two. If you get stuck I don’t mind helping but I thought this is the best way to get you started.

You might be interested in

14. Molly puts two shapes together to make a new shape. The new shape has 6 sides and 6 vertices. Which two shapes did she use?

olganol [36]

She used two trapezoids.

She put one trapezoid on top and one on the bottom.

The shape she ends up making is a hexagon.

3 0

3 years ago

Mrs. Thornton is teaching a 5th grade class. She is standing 15 meters in front of Hillary. Avisha is sitting 8 meters to Hillar

kykrilka [37]

15+8=23, i think is the answer

5 0

3 years ago

Q: Out of 20 attempts a basketball player scored 8 times. What percent of

zhenek [66]

I think it’s b but i’m not sure

6 0

3 years ago

Read 2 more answers

What value is equivalent to 23 · 34?

In-s [12.5K]

Answer:

782

Step-by-step explanation:

You have to multiplied 23 and 34

23 X 34 = 782

4 0

3 years ago

Read 2 more answers

What are the roots of the polynomial equation x3 - 6x = 3x2 - 8? Use a graphing calculator and a system of equations.

makkiz [27]

Answer:

d : 4, 1, -2.

Step-by-step explanation:

There's no need for a calculator in my opinion because we can use the rational root theorem which states that the equations of this form:

$\sum_{i=0}^{n} a_{i}x^i = 0 = a_{0} + a_{1}x + a_2x^2 + ... + a_nx^n = 0\\ a_i \in \mathbb{Z}$

Have rational roots, than the roots are of the form: $\frac{k \cdot a_o}{p \cdot a_n}, \ where \ k, p \in \mathbb{Z}.$

Rewriting the equation we have:

$x^3 - 3x^2 -6x + 8 = 0\\By \ our\ previous \ claim \ x \in D_8 \ where \ D_n \ the \ set \ of \ divisors \ of \ n.\\D_8 = \{\pm 1, \pm 2, \pm 4, \pm 8\} \\We \ plug \ in \ some \ number from \ D_8.\\Letting \ x = -2;\\(-2)^3 -3(-2)^2 -6(-2) + 8 = 0 \ \ \ Thus \ x+2 \ is \ a \ factor.\\We \ can \ now \ simplify \ the \ eq. \ using \ Polynomial \ Long \ Division.$

$x^3 -3x^2 -6x + 8 = (x+2)(Q(x))\\To \ find \ Q \ we \ divide \ the \ original \ equation \ by \ x + 2; which \ yields:\\Q(x) = x^2 - 5x + 4.\\So \ we \ can \ use \ the \ quadratic \ formula \ to \ find \ the \ roots:\\x = 4,x = 1. \\Thus \ x \in \{-2, 1, 4\}. \ \ These \ are \ all \ the \ roots.$

5 0

4 years ago