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aleksandrvk [35]

2 years ago

6

Solve for n. 15 = 0.5n

1 answer:

murzikaleks [220]2 years ago

6 0

15 = 0.5n

15/0.5 = n

15/0.5 = 30

n = 30.

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Please answer this question right now

Westkost [7]

Answer:

6) c

7) d

Step-by-step explanation:

6) 25 + 15= 40

40+35=75

Php. 75

7) 29•3=87

150-87=63

Php. 63

Hope this helps!

6 0

2 years ago

A math test has 40 questions. A student answers 85% of the questions correctly. How many questions did the student answer correc

Novosadov [1.4K]

Answer:

34

Step-by-step explanation:

40 * 0.85 = 34

34 questions answered correctly

6 0

1 year ago

Read 2 more answers

A day can be evenly divided into 86,400 periods of 1 second; 43,200 periods of each 2 seconds; or in many other ways. In total,

Leni [432]

Answer:

96 ways

Step-by-step explanation:

Given

$Day = 86400\ seconds$

Required

Ways to divide it into period of seconds

What this question implies is to determine the total number of factors of 86400

To start with, we determine the prime factorization of 86400

To do this, we continually divide 86400 by 2; when it can not be further divided, we divide by 3, then 7, then 11...

$86400/2 = 43200$

$43200/2=21600$

$21600/2=10800$

$10800/2=5400$

$5400/2= 2700$

$2700/2 = 1350$

$1350/2=675$

$675/3=225$

$225/3=75$

$75/3=25$

$25/5=5$

$5/5 = 1$

This implies that:

$86400 = 2^7 * 3^3 * 5^2$

The number of factors d is the solved by:

$d = (a+1)*(b+1) *(c+1)$

Where

$n = 2^a * 3^b * 5^c$

By comparison:

$a = 7$

$b = 3$

$c=2$

So:

$d = (7+1)*(3+1) *(2+1)$

$d = 8*4 *3$

$d = 96$

<em>Hence, there are 96 total ways</em>

7 0

2 years ago

Please answer asap this is 7th grade math<br> find the area

Rainbow [258]

25pi

pi x 10^2 is 100pi

Divide by four = 25pi

7 0

2 years ago

Read 2 more answers

In Exercises 10 and 11, points B and D are points of tangency. Find the value(s) of x.<br>

Lemur [1.5K]

In both cases,

$AB^2=AD^2$

(as a consequence of the interesecting secant-tangent theorem)

So we have

10.

$(4x+7)^2=(6x-3)^2$

$16x^2+56x+49=36x^2-36x+9$

$20x^2-92x-40=0$

$5x^2-23x-10=0$

$(5x+2)(x-5)=0\implies\boxed{x=5}$

(omit the negative solution because that would make at least one of AB or AD have negative length)

11.

$(4x^2-18x-10)^2=(x^2+x+4)^2$

$16x^4-144x^3+244x^2+360x+100=x^4+2x^3+9x^2+8x+16$

$15x^4-146x^3+235x^2+352x+84=0$

$(x-7)(3x+2)(5x^2-17x-6)=0\implies\boxed{x=-\dfrac23\text{ or }x=7}$

(again, omit the solutions that would give a negative length for either AB or AD)

7 0

3 years ago