Write a system of linear equations that has the ordered pair as its solution.

HURRY I NEED THIS BEFORE 12

I am Lyosha [343]

X=8 because add 11 to both sides and it becomes 28=3.5x and divide both sides by 3.5 so x=8

5 0

2 years ago

Which is bigger 5/8 or 7/12

gladu [14]

5/8=0.625

7/12=0.583......

5/8 is bigger as ive put both into a decimal as it is easier to tell

4 0

3 years ago

Read 2 more answers

I need helps pls :)

nadya68 [22]

In regular polygons, if the number of sides, n, is odd, the lines of symmetry will pass through a vertex and the half of the opposite side.
Step-by-step explanation:
This is equivalent to say that a perpendicular bisector of a side of the regular polygon bisects the angle of the opposite vertex to the side, and divide the polygon into two symmetrical parts.

4 0

2 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

2 years ago

How many distinct products can be formed using two different integers from the given set: {–6, –5, –4, –3, –2, –1, 0, 1, 2, 3, 4

zhannawk [14.2K]

Number of distinct products that can be formed is 144

<h3>Permutation</h3>

Since we need to multiply two different integers to be selected from the set which contains a total of 12 integers. This is a permutation problem since we require distinct integers.

Now, for the first integer to be selected for the product, since we have 12 integers, it is to be arranged in 1 way. So, the permutation is ¹²P₁ = 12

For the second integer, we also have 12 integers to choose from to be arranged in 1 way. So, the permutation is ¹²P₁ = 12.

<h3>Number of distinct products</h3>

So, the number of distinct products that can be formed from these two integers are ¹²P₁ × ¹²P₁ = 12 × 12 = 144

So, the number of distinct products that can be formed is 144

Learn more about permutation here:

brainly.com/question/25925367

6 0

2 years ago