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lilavasa [31]
3 years ago
5

CAN SOMEONE PLEASE HELP I WILL GIVE BRAINLIEST TO HOWEVER ANSWERS FIRST!!

Mathematics
1 answer:
igomit [66]3 years ago
3 0
Answer: y+5= -1/7(x-7)

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What is 7x^5-33x^4-4x^2+3x+52=0
lorasvet [3.4K]
7x^5-33x^4-4x^2+3x+52=

ANSWER: 7x^5 -33x^4 -4x^2+3x+52  (because there are no like terms)
                  
5 0
3 years ago
Which of the following equations has exactly one solution?
shusha [124]

Answer:

A

Step-by-step explanation:

5a+4=9a+3-4a

5a-9a+4a=3-4

a=-1

6 0
2 years ago
Tanya’s rotation maps point K(24, –15) to K’(–15, –24). Which describes the rotation?
harina [27]
90 degree rotation clockwise : (a,b) --> (b,-a)
so... (24,-15) --> (-15,-24) is a 90 degree clockwise rotation
4 0
3 years ago
Read 2 more answers
The ages of George and Genemy are in ratio 5:7 . Four years from now the ratio of their ages will be 3:4 .Find their present age
DedPeter [7]

Answer:

Below in bold.

Step-by-step explanation:

Let x be George's age and y be Genemy's age.

x/y = 5/7

(x + 4)/ (y + 4) = 3/4

From equation 1, y = 7x/5 = 1.4x so substituting in equation 2 :

(x + 4)/ (1.4x + 4) = 3/4

3(1.4x + 4) = 4(x + 4)

4.2x + 12 = 4x + 16

0.2x = 16 -12 = 4

x = 4/0.2 = 20 = George's age.

So y = 1.4x = 1.4 * 20 = 28 = Genemy's age.

3 0
3 years ago
Read 2 more answers
A fair coin is tossed three times and the events A, B, and C are defined as follows: A:{ At least one head is observed } B:{ At
kicyunya [14]

Answer:

(a) 1/2

(b) 1/2

(c) 1/8

Step-by-step explanation:

Since, when a fair coin is tossed three times,

The the total number of possible outcomes

n(S) = 2 × 2 × 2

= 8 { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT },

Here, B : { At least two heads are observed } ,

⇒ B = {HHH, HHT, HTH, THH},

⇒ n(B) = 4,

Since,

\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}

(a) So, the probability of B,

P(B) =\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}

(b) A : { At least one head is observed },

⇒ A = {HHH, HHT, HTH, THH, HTT, THT, TTH},

∵ A ∩ B = {HHH, HHT, HTH, THH},

n(A∩ B) = 4,

\implies P(A\cap B) = \frac{n(A\cap B)}{n(S)} = \frac{4}{8}=\frac{1}{2}

(c) C: { The number of heads observed is odd },

⇒ C = { HHH, HTT, THT, TTH},

∵ A ∩ B ∩ C = {HHH},

⇒ n(A ∩ B ∩ C) = 1,

\implies P(A\cap B\cap C)=\frac{1}{8}

7 0
3 years ago
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