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3 years ago

G − 334 = –234 solve for g

Mathematics

2 answers:

s344n2d4d5 [400]3 years ago

6 0

Answer:

100

Step-by-step explanation:

g-334= -234

g-334+ 334= -234 + 334

g=100

Harrizon [31]3 years ago

4 0

Answer: 100

Step-by-step explanation: To find g, add 334 to both sides. -234 + 334 is 100.

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Write 3,420,000 in scientific notation.

olga55 [171]

3.420 times 10 to the 6th power will equal 3,420,000

5 0

4 years ago

Read 2 more answers

If Sam and Sally invested the same total amount at the end of three years, the amount Sam invested the first year is $ and the a

Vlada [557]

Answer:

The amount Sam invested the first year = $2000

The amount Sally invested the last year = $1900

Complete question related to this was found at brainly (ID 4527784):

For three consecutive years, Sam invested some money at the start of the year. The first year, he invested x dollars. The second year, he invested $2,000 less than 5/2 times the amount he invested the first year. The third year, he invested $1,000 more than 1/5 of the amount he invested the first year.

During the same three years, Sally also invested some money at the start of every year. The first year, she invested $1,000 less than 3/2 times the amount Sam invested the first year. The second year, she invested $1,500 less than 2 times the amount Sam invested the first year. The third year, she invested $1,400 more than 1/4 of the amount Sam invested the first year.

If Sam and Sally invested the same total amount at the end of three years, the amount Sam invested the first year is $ and the amount Sally invested the last year is $ .

Step-by-step explanation:

First we would represent the information given with mathematical expressions.

Sam investment for 3 consecutive years:

Year 1 = x dollars

Year 2 = $2,000 less than 5/2 times the amount he invested the first year

Year 2 = (5/2)(x) - 2000

Year 3 = $1,000 more than 1/5 of the amount he invested the first year

Year 3 = (1/5)(x) + 1000

Sally investment for 3 consecutive years:

Year 1 = $1,000 less than 3/2 times the amount Sam invested the first year

Year 1 = (3/2)(x) - 1000

Year 2 = $1,500 less than 2 times the amount Sam invested the first year

Year 2 = 2x - 1500

Year 3 = $1,400 more than 1/4 of the amount Sam invested the first year.

Year 3 = (1/4)(x) + 1400

Since Sam and Sally invested the same total amount at the end of three years, we would equate their sum:

Sum of Sam investment for the 3years = x + (5/2)(x) - 2000 + (1/5)(x) + 1000

= x + 5x/2 -2000 + x/5 + 1000

= (10x+25x+2x)/10 - 1000

= 37x/10 - 1000

Sum of Sally investment for the 3years = (3/2)(x) - 1000 + 2x - 1500 + (1/4)(x) + 1400

= 3x/2 - 1000 + 2x -1500 + x/4 + 1400

= (6x+8x+x)/4 - 1100

= 15x/4 - 1100

37x/10 - 1000 = 15x/4 - 1100

37x/10 - 15x/4 = -100

(148x - 150x)/40 = -100

-2x = -4000

x = 2000

Therefore the amount Sam invested the first year = x = $2000

The amount Sally invested the last year (3rd year) = (1/4)(x) + 1400

(1/4)(2000) + 1400 = 500+1400 = 1900

The amount Sally invested the last year = $1900

8 0

3 years ago

Heeeeeeeeeeeelp !!!!!!!!!!!!

Marysya12 [62]

Answer:

Step-by-step explanation:

6x2 + 5x - 4

6x2 + 8x - 3x - 4

2x(3x + 4) - 1(3x + 4)

(2x - 1) (3x + 4)

These are the factors

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3 years ago

From a shipment of 65 transistors, 6 of which are defective, a sample of 5 transistors is selected at random.

Vladimir79 [104]

Answer:

a) 8259888

b) 34220

c) 45057474

Step-by-step explanation:

Given,

The total number of transistor = 65,

In which, the defective transistor = 6,

So, the number of non defective transistor = 65 - 6 = 59,

Since, out of these transistor 5 are selected,

a) Thus, the number of ways = the total possible combination of 5 transistors = ${65}C_ 5$

$=\frac{65!}{(65-5)!5!}$

$=8259888$

b) The number of samples that contains exactly 3 defective transistors = the possible combination of exactly 3 defective transistors = ${6}C_3\times {59}C_2$