Point E is located at (–1, 1). Which of the following choices contain end points of the line segment that has E as its midpoint?

Can someone pls pls pls pls pls pls answer these question

Lisa [10]

Answer:

Decay

Step-by-step explanation:

6 0

3 years ago

Can someone help pls

lora16 [44]

I got 7.1 x 10^(-11) but it's been a few years so I could've done it wrong

5 0

4 years ago

In ΔGHI, the measure of ∠I=90°, the measure of ∠G=46°, and IG = 87 feet. Find the length of GH to the nearest tenth of a foot.

sergij07 [2.7K]

Answer:

See attached

I got an answer of 125.2411251548

or to the nearest tenth 125.2

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Mario took clothes to the cleaners three times last month. first, he brought 3 shorts and 1pair of slacks and $10.96. then, he b

valina [46]

Let us say that:

x = cost of shorts

y = cost of slacks

z = cost of sports coat

From the given statements, we can create the following equations:

eqtn 1: 3 x + y = 10.96

eqtn 2: 7 x + 2 y + z = 30.40

eqtn 3: 4 x + z = 14.45

Rewrite eqtn 1 explicit to y:

y = 10.96 – 3 x --> eqtn 4

Rewrite eqtn 3 explicit to z:

z = 14.45 – 4x --> eqtn 5

Plug in the values of y (eqtn 4) and z (eqtn 5) to eqtn 2:

7 x + 2 (10.96 – 3 x) + 14.45 – 4 x = 30.40

7 x + 21.92 – 6 x + 14.45 – 4 x = 30.40

- 3 x + 36.37 = 30.40

- 3 x = - 5.97

x = 1.99

Eqtn 4:

y = 10.96 – 3 x

y = 10.96 – 3 (1.99)

y = 4.99

Eqtn 5:

z = 14.45 – 4x

z = 14.45 – 4 (1.99)

z = 6.49

Therefore each shirt cost $1.99, each slacks cost $4.99, and each sports coat cost $6.49.

7 0

4 years 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