All categories

3 years ago

X-12y=-210 and x-6=90

Mathematics

1 answer:

m_a_m_a [10]3 years ago

7 0

Answer:

x = -30

y = -20

Step-by-step explanation:

x - 12y = 210..... (I)

x - 6y = 90..... (II)

Subtract equation (II) from (I)

- 12y - (-6y) = 210 - 90

-12y + 6y = 120

-6y = 120

Divide through by -6y

Therefore, y = -20

Substitute y = -20 in equation (II)

x -6 (-20) = 90

x + 120 = 90

x = 90 - 120

x = -30

Hope that helped

You might be interested in

Let H be a subgroup of a group G. We call H characteristic in G if for any automorphism σ∈Aut(G) of G, we have σ(H)=H.

choli [55]

Answer:Problem 1. Let G be a group and let H, K be two subgroups of G. Dene the set HK = {hk : h ∈ H,k ∈ K}.

a) Prove that if both H and K are normal then H ∩ K is also a normal subgroup of G.

b) Prove that if H is normal then H ∩ K is a normal subgroup of K.

c) Prove that if H is normal then HK = KH and HK is a subgroup of G.

d) Prove that if both H and K are normal then HK is a normal subgroup of G.

e) What is HK when G = D16, H = {I,S}, K = {I,T2,T4,T6}? Can you give geometric description of HK?

Solution: a) We know that H ∩ K is a subgroup (Problem 3a) of homework 33). In order to prove that it is a normal subgroup let g ∈ G and h ∈ H ∩ K. Thus h ∈ H and h ∈ K. Since both H and K are normal, we have ghg−1 ∈ H and ghg−1 ∈ K. Consequently, ghg−1 ∈ H ∩ K, which proves that H ∩ K is a normal subgroup.

b) Suppose that H G. Let K ∈ k and h ∈ H ∩ K. Then khk−1 ∈ H (since H is normal in G) and khk−1 ∈ K (since both h and k are in K), so khk−1 ∈ H ∩ K. This proves that H ∩ K K.

c) Let x ∈ HK. Then x = hk for some h ∈ H and k ∈ K. Note that x = hk = k(k−1hk). Since k ∈ K and k−1hk ∈ H (here we use the assumption that H G), we see that x ∈ KH. This shows that HK ⊆ KH. To see the opposite inclusion, consider y ∈ KH, so y = kh for some h ∈ H and k ∈ K. Thus y = (khk−1)k ∈ HK, which proves that KH ⊆ HK and therefoere HK = KH. To prove that HK is a subgroup note that e = e · e ∈ HK. If a,b ∈ HK then a = hk and b = h1k1 for some h,h1 ∈ H and k,k1 ∈ K. Thus ab = hkh1k1. Since HK = KH and kh1 ∈ KH, we have kh1 = h2k2 for some k2 ∈ K, h2 ∈ H. Consequently,

ab = h(kh1)k1 = h(h2k2)k1 = (hh2)(k2k1) ∈ HK

(since hh2 ∈ H and k2k1 ∈ K). Thus HK is closed under multiplication. Finally,

Step-by-step explanation:

6 0

3 years ago

Solve the right triangle ABC, with C = 90°.

Yanka [14]

Answers are as follows
B = 51.5
a = 34.6
c = 55.6

6 0

2 years ago

BRAINLIEST-<br> Find the volume of the figure

EleoNora [17]

Answer:

4080cm²

Step-by-step explanation:

3 0

3 years ago

Mary and Jane both took PSY230 last semester but they were in two different sections taught by different professors. Mary's clas

Greeley [361]

Answer:

Due to the higher z-score, Jane did better in class.

Step-by-step explanation:

Z-score:

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

Whoever's grade had the better z-score did better in class.

Mary's:

Mean of 85, standard deviation of 5, grade of 88. This means that $\mu = 85, \sigma = 5, X = 88$ . So