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3 years ago

983 divided by 6 in base time blocks

Mathematics

2 answers:

castortr0y [4]3 years ago

4 0

The answer will probably be 163.83

dusya [7]3 years ago

3 0

The answer is 163.83

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Students are given 3 minutes to complete each multiple-choice question on a test and 8 minutes for each free-response question.

SpyIntel [72]

Let

x------> the number of multiple choice question

y------> the number of free response question

we know that

$3x+8y=55$ -----> equation A

$x+y=15$

$x=15-y$ -----> equation B

Substitute equation B in equation A

$3[15-y]+8y=55$

$45-3y+8y=55$

$5y=55-45$

$5y=10$

$y=2$

Find the value of x

$x=15-y$

$x=15-2=13$

therefore

<u>the answer is</u>

the number of multiple choice question are $13$

the number of free response question are $2$

5 0

3 years ago

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A pack of five basic solid T-shirts costs $29.95 in a store. The sales tax where they are sold is 7.75%. What is the unit cost o

Anna71 [15]

Answer:

6.45

Step-by-step explanation:

29.95+7.75%=32.27

32.27/5

6.45

3 0

3 years ago

Read 2 more answers

I need help ☝️ urgent please hurry up man

muminat

Answer:

J) 110.88 mm^2

Step-by-step explanation:

height =6.93x2=13.86

Area=13.86x16x0.5=110.88

7 0

3 years ago

Find the midpoint of points A (5,-6) and B(2,0) graphically

netineya [11]

I think its (3.5, -3)

8 0

2 years ago

Anyone knows how to do questions 7 and 8? 15 pts!!

Oksi-84 [34.3K]

7) Certainly there is a typo in the statement, just see that the expression of item (ii) is different from that of item (i). Probably the correct expression is: $2x^2-4x+5$ . With this consideration, we can continue.

(i) Let E the expression that we are analyzing:

$E=2x^2-4x+5\\\\ E=2x^2-4x+2-2+5\\\\ E=2(x^2-2x+1)-2+5\\\\ E=2(x-1)^2+3$

Since (x-1)² is a perfect square, it is a positive number. So, E is a result of a sum of two positive numbers, 2(x-1)² and 3. Hence, E is a positive number, too.

(ii) Manipulating the expression:

$2x^2+5=4x\\\\ 2x^2-4x+5=0$

So, it's the case when E=0. However, E is always a positive number. Then, there is no real number x that satisfies the expression.

8) Let E the expression that we want to calculate:

$E=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)+1\\\\ E-1=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)$

Multiplying by (2-1) in the both sides:

$(2-1)(E-1)=(2-1)(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2-1)(2+1)}_{2^2-1}(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2^2-1)(2^2+1)}_{2^4-1}(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ ...$ Repeating the process, we obtain: $...\\\\ E-1=(2^{32}-1)(2^{32}+1)\\\\ E-1=2^{64}-1\\\\ \boxed{E=2^{64}}$

3 0

3 years ago

Read 2 more answers