Could someone with this

Geometry. im giving 25 points! Please help me.

nata0808 [166]

Message me and I can help you

6 0

3 years ago

Can someone answer the question ??

Temka [501]

Answer:
Faris' arrival time = 3:30 pm + 3 h 54 mnt = 7:24 pm
Step-by-step explanation:
Driving distance between Manchester to London = 195 miles
Coach left Manchester = 3.30 pm
Average speed of coach = 50 mph
Total time taken by the coach to reach the Manchester to London = 195÷50 = 3.9 h
0.9 h = 0.9×60 mnt = 54 mnt
Now,
3.9 h = 3 h 54 mnt
Hence Faris' arrival time in London = 3.30 pm + 3 h 54 mnt = 7.24 pm

5 0

3 years ago

Some one help me i lost with this thing

egoroff_w [7]

Answer:

Im in middle school

Step-by-step explanation: I have no clue how to help u and bc I have no idea how brainy works -_-

4 0

3 years ago

Jennifer had 7/8 foot board she cut off 1/4 foot peice that was for a project.in feet how much board was left?

guapka [62]

$\frac{5}{8} \text{ feet board is left }$

<em><u>Solution:</u></em>

Given that, Jennifer had 7/8 foot board she cut off 1/4 foot peice that was for a project

To find: Board that was left

From given,

$\text{Total length of board } =\frac{7}{8} \text{ feet }$

$\text{Foot that was cut off } = \frac{1}{4} \text{ feet }$

Remaining feet of board is found by finding the difference between total length of board and foot that was cut off

$\text{Reamining board } = \text{Total length of board - feet of board that was cut off}\\$

Substituting the values we get,

$\text{Remaining board } = \frac{7}{8} - \frac{1}{4}\\\\\text{Remaining board } =\frac{7}{8} - \frac{1 \times 2}{4 \times 2}\\\\\text{Remaining board } =\frac{7}{8} -\frac{2}{8} = \frac{7-2}{8}\\\\\text{Remaining board } =\frac{5}{8}$

Thus $\frac{5}{8}$ feet board is left

5 0

3 years ago

Using the pattern, give the coefficients of (x + y)^5 and (x + y)^6

musickatia [10]

Answer:

$(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$

Step-by-step explanation:

In order to find the values of $(x+y)^5$ and $(x+y)^6$ , you need to apply the binomial theorem (high-level math you most likely don't need to worry about, it's easier than multiplying all the binomials together).

$(x+y)^5$ = $\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$ = $\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$ = $x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$ .

$(x+y)^6$ = $\sum _{i=0}^6\binom{6}{i}x^{\left(6-i\right)}y^i$ = $\frac{6!}{0!\left(6-0\right)!}x^6y^0+\frac{6!}{1!\left(6-1\right)!}x^5y^1+\frac{6!}{2!\left(6-2\right)!}x^4y^2+\frac{6!}{3!\left(6-3\right)!}x^3y^3+\frac{6!}{4!\left(6-4\right)!}x^2y^4+\frac{6!}{5!\left(6-5\right)!}x^1y^5+\frac{6!}{6!\left(6-6\right)!}x^0y^6$ = $x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$ .

5 0

3 years ago