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Romashka-Z-Leto [24]
3 years ago
11

Which are perfect squares?

Mathematics
2 answers:
Digiron [165]3 years ago
8 0

Answer:

4

225

9

ep-by-step explanation:

grandymaker [24]3 years ago
3 0

Answer:

Step-by-step explanation:

A perfect square is an integer that is the square of another integer.

49 = 7*7 / 7 squared

9 = 3*3 / 3 squared

225 = 15*15 / 15 squared

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Solve for j.<br> 2j - k= 5
Andreas93 [3]

Answer:

j=1/2k+5/2

Step-by-step explanation:

2j−k=5

Step 1: Add k to both sides.

2j−k+k=5+k

2j=k+5

Step 2: Divide both sides by 2.

2j/2=k+5/2

j=1/2k+5/2

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What is the value of 7-3+(6-2•4)
defon

Answer:

2

Step-by-step explanation:

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Which exponential function has an initial value of 2?<br><br> f(x) = 2(3x)<br><br> f(x) = 3(2x)
Sidana [21]
<span>f(x) = 2(3x)
Exponential functions represent the initial value outside of the parentheses so if 2 is the initial value it has to be on the outside of the parentheses. 
Exponential growth formula.
</span>y=a(1+r)^{x}
<span>a represents the initial value.</span>
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What is the value of the function when x = −1?<br><br> Is my answer right?
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The answer would be negative 2
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Please dont ignore, Need help!!! Use the law of sines/cosines to find..
Ket [755]

Answer:

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Step-by-step explanation:

<h3>16</h3>

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

  • \sin{A} = \sin{103\textdegree{}},
  • The opposite side of angle A a = BC = 26,
  • The angle C is to be found, and
  • The length of the side opposite to angle C c = AB = 6.

\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}.

\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}.

\displaystyle C = \sin^{-1}{(\sin{C}}) = \sin^{-1}{\left(\frac{c}{a}\cdot \sin{A}\right)} = \sin^{-1}{\left(\frac{6}{26}\times \sin{103\textdegree}}\right)} = 13.0\textdegree{}.

Note that the inverse sine function here \sin^{-1}() is also known as arcsin.

<h3>17</h3>

By the law of cosine,

c^{2} = a^{2} + b^{2} - 2\;a\cdot b\cdot \cos{C},

where

  • a, b, and c are the lengths of sides of triangle ABC, and
  • \cos{C} is the cosine of angle C.

For triangle ABC:

  • b = 21,
  • c = 30,
  • The length of a (segment BC) is to be found, and
  • The cosine of angle A is \cos{123\textdegree}.

Therefore, replace C in the equation with A, and the law of cosine will become:

a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}.

\displaystyle \begin{aligned}a &= \sqrt{b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}}\\&=\sqrt{21^{2} + 30^{2} - 2\times 21\times 30 \times \cos{123\textdegree}}\\&=45.0 \end{aligned}.

<h3>18</h3>

For triangle ABC:

  • a = 14,
  • b = 9,
  • c = 6, and
  • Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}.

\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}.

\displaystyle B = \cos^{-1}{\left(\frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}\right)} = \cos^{-1}{\left(\frac{14^{2} + 6^{2} - 9^{2}}{2\times 14\times 6}\right)} = 26.0\textdegree.

<h3>15</h3>

For triangle DEF:

  • The length of segment DF is to be found,
  • The length of segment EF is 9,
  • The sine of angle E is \sin{64\textdegree}}, and
  • The sine of angle D is \sin{39\textdegree}.

Apply the law of sine:

\displaystyle \frac{DF}{EF} = \frac{\sin{E}}{\sin{D}}

\displaystyle DF = \frac{\sin{E}}{\sin{D}}\cdot EF = \frac{\sin{64\textdegree}}{39\textdegree} \times 9 = 12.9.

7 0
3 years ago
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