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kherson [118]
4 years ago
5

there are 320 seniors at strong field high school. if 90% of them go expect to go to college, how many of the seniors expect to

go to college?
Mathematics
2 answers:
weqwewe [10]4 years ago
8 0
288. to figure this out I set up a proporton, 90 (percent) over 100 ( percentages are always out of 100) and x ( the variable, what's ere trying to figure out) over 320 (since we are trying to find what over 320 is equal to 90 out of 100.
\frac{90}{100}  =  \frac{x}{320}
after cross multiplying, you find that 90/100 is equal to 288/320
deff fn [24]4 years ago
7 0

\bf a\%~of~b \implies \cfrac{ a}{100}\cdot b~\hfill
\textit{90\% of 320 is }\cfrac{90}{100}\cdot 320

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Which statement about the following sequence is true? -10, -3, -10, -3, -10, ... Question 3 options: It is arithmetic. It is geo
Novosadov [1.4K]

Answer:

What is the value of a2 in the sequence?

2, 5, 10, 17, ... is 2

What is the value of a3 in the sequence?

−4, −2, −1/2, −1/4, ... is ​ −1/2 ​

Which answer describes the type of sequence?

3, 9, 15, 21, ...is arithmetic with d=6

Which answer describes the type of sequence?

4, 9, 13, 22, ...

is neither arithmetic nor geometric

Which answer describes the type of sequence?

200, 40, 8, 1.6, ...

is geometric with q=1/5

 Have fun

7 0
3 years ago
Read 2 more answers
How do you solve this problem? what are the steps? what to do?
adell [148]
Micho tito sancho peludo negro y cabezon

3 0
3 years ago
Pls help! <br> and a reason.
Rasek [7]

Answer:

\angle BDA = 72^{\circ}

\angle DAB = 36^{\circ}

\angle BDC = 108^{\circ}

\angle BCD = 36^{\circ}

\angle DBC = 36^{\circ}

Step-by-step explanation:

Given

\angle ABD = 72^{\circ}  

Required

Determine the missing angles

Since AB = AD, then:

\angle ABD = \angle BDA --- Base angles of an isosceles triangle

Hence:

\angle ABD = \angle BDA = 72^{\circ}

\angle ABD + \angle BDA + \angle DAB = 180^{\circ} --- angles in a triangle

72^{\circ} + 72^{\circ} + \angle DAB = 180^{\circ}

144^{\circ} + \angle DAB = 180^{\circ}

Collect Like Terms

\angle DAB = 180^{\circ} - 144^{\circ}

\angle DAB = 36^{\circ}

\angle BDA + \angle BDC = 180^{\circ} --- angle on a straight line

72^{\circ} + \angle BDC = 180^{\circ}

\angle BDC = 180^{\circ} - 72^{\circ}

\angle BDC = 108^{\circ}

Since ABC is isosceles, then:

\angle DAB = \angle BCD = 36^{\circ} --- base angle of isosceles

\angle BCD = 36^{\circ}

Lastly:

\angle BCD + \angle BDC + \angle DBC = 180^{\circ}

36^{\circ} + 108^{\circ} + \angle DBC = 180^{\circ}

\angle DBC = 180^{\circ} - 36^{\circ} - 108^{\circ}

\angle DBC = 36^{\circ}

<em>See attachment</em>

3 0
3 years ago
Please help solve this question
sineoko [7]
(-2x^3y^4z^0)^4

<span><span><span>(−2<span>x^<span><span>​3</span><span>​​^</span></span></span><span>y<span><span>​4</span><span>​​</span></span></span>×^1)</span><span><span>​4</span><span>​​
</span></span></span><span>
(-2x^3y^4)^4

(2x^3y^4)^4

2^4(x^3)^4(y^4)^4

16(x^3)^4(y^4)^4

16x^12(y^4)^4</span></span>
8 0
3 years ago
Read 2 more answers
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