I need helppppp asappp!!

Amir has $200 budget to spend on a graduation party for his son he already has purchased $122 worth of drinks and party supplies

Bas_tet [7]

Answer: C, D, and E

Step-by-step explanation:

200-122=78 remaining

A: 7*11.75= 82.25

B: 10*11.75= 117.50

C: 9*6.80=61.20

D: 11*6.80=74.80

E: 70*1.10=77.00

F: 71*1.10=78.10

Only C, D and E are below the dollars remaining in the budget.

5 0

3 years ago

Anthony travels from Page Az to Flagstaff Az at a rate of 60 miles per hour in 2.5 hours. How far is Flagstaff from Page?( Show

DerKrebs [107]

Answer:

150 miles

Step-by-step explanation: The distance from Page to Flagstaff can be shown with the equation d = r * t, or distance = rate * time. In this case, the rate is 60 mph, and the rate is 2.5 h. Plugging that into the equation you get d = 2.5(60) = 150, giving you your distance.

4 0

3 years ago

21 less than tree-fifths of a number

MissTica

Answer:

$\frac{3}{5}x - 21$

Step-by-step explanation

Three-fifths of a number means $\frac{3}{5}x$ and 21 less than that number means $\frac{3}{5} x - 21$ .

3 0

3 years ago

34. For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.

skad [1K]

Answer:

The linear equation for the line which passes through the points given as (-1,4) and (5,2), is written in the point-slope form as $y=\frac{1}{3} x-\frac{13}{3}$ .

Step-by-step explanation:

A condition is given that a line passes through the points whose coordinates are (-1,4) and (5,2).

It is asked to find the linear equation which satisfies the given condition.

Step 1 of 3

Determine the slope of the line.

The points through which the line passes are given as (-1,4) and (5,2). Next, the formula for the slope is given as,

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Substitute 2&4 for $y_{2}$ and $y_{1}$ respectively, and $5 \&-1$ for $x_{2}$ and $x_{1}$ respectively in the above formula. Then simplify to get the slope as follows,

$m=\frac{2-4}{5-(-1)}$\\ $m=\frac{-2}{6}$\\ $m=-\frac{1}{3}$$

Step 2 of 3

Write the linear equation in point-slope form.

A linear equation in point slope form is given as,

$y-y_{1}=m\left(x-x_{1}\right)$

Substitute $-\frac{1}{3}$ for m,-1 for $x_{1}$ , and 4 for $y_{1}$ in the above equation and simplify using the distributive property as follows,

$y-4=-\frac{1}{3}(x-(-1))$\\ $y-4=-\frac{1}{3}(x+1)$\\ $y-4=-\frac{1}{3} x-\frac{1}{3}$$

Step 3 of 3

Simplify the equation further.

Add 4 on each side of the equation $y-4=\frac{1}{3} x-\frac{1}{3}$ , and simplify as follows,

$y-4+4=\frac{1}{3} x-\frac{1}{3}+4$\\ $y=\frac{1}{3} x-\frac{1+12}{3}$\\ $y=\frac{1}{3} x-\frac{13}{3}$$

This is the required linear equation.

5 0

1 year ago

The n term of a geometric sequence is denoted by Tn and the sum of the first n terms is denoted by Sn.Given T6-T4=5/2 and S5-S3=

Leno4ka [110]

1 step: $S_{5}=T_{1}+T_{2}+T_{3}+T_{4}+T_{5}$ , $S_{3}=T_{1}+T_{2}+T_{3}$ , then
$S_{5}-S_{3}=T_{4}+T_{5}=5$ .

2 step: $T_{n}=T_{1}*q^{n-1}$ , then
$T_{6}=T_{1}*q^{5}$
$T_{5}=T_{1}*q^{4}$
$T_{4}=T_{1}*q^{3}$
$T_{3}=T_{1}*q^{2}$
and $\left \{ {{T_{6}-T_{4}= \frac{5}{2} } \atop {T_{5}+T_{4}=5}} \right.$ will have form $\left \{ {{T_1*q^{5}-T_{1}*q^{3}= \frac{5}{2} } \atop {T_{1}*q^{4}+T_{1}*q^{3}=5} \right.$ .

3 step: Solve this system $\left \{ {{T_1*q^{3}*(q^{2}-1)= \frac{5}{2} } \atop {T_{1}*q^{3}*(q+1)=5} \right.$ and dividing first equation on second we obtain $\frac{q^{2}-1}{q+1}= \frac{ \frac{5}{2} }{5}$ . So, $\frac{(q-1)(q+1)}{q+1} = \frac{1}{2}$ and $q-1= \frac{1}{2}$ , $q= \frac{3}{2}$ - the common ratio.

4 step: Insert $q= \frac{3}{2}$ into equation $T_{1}*q^{3}*(q+1)=5$ and obtain $T_{1}* \frac{27}{8}*( \frac{3}{2}+1 ) =5$ , from where $T_{1}= \frac{16}{27}$ .

5 0

3 years ago