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Maslowich
3 years ago
15

Which phrase is a description of s - 12?

Mathematics
1 answer:
scZoUnD [109]3 years ago
5 0

Answer:

12 less than a number s

Step-by-step explanation:

less than means after

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Answer:5

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Plz help this is due soon​
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put a dot on the following points ( -1, -3) (0, -2) (1, -1) and (2, 0)

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2 years ago
Each of four friends orders a sweatshirt from a catalog. There are 14 colors of sweatshirts, 7 of which are all cotton and 7 of
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Multiply these together: (7/26)*(6/15)*(5/14)*(4/13)
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3 0
3 years ago
Read 2 more answers
The diagram shows a 9cm x 7cm rectangle-based pyramid. all the diagonal sides - TA, TB, TC, AND TD- are length 12cm. M is the mi
lakkis [162]

Answer:

∠TAC is approximately equal to 61.6°

Step-by-step explanation:

The given parameters for the pyramid are;

The dimension for the rectangular base are; Length = 9 cm, width = 7 cm

The length of the diagonal sides, TA, TB, TC, and TD = 12 cm each

The midpoint of the rectangular base = Point M

The diagonal AC = AM + MC

AM = MC as given M is the midpoint of the rectangular base

∴ AC = AM + MC = 2·AM

By Pythagoras' theorem, AC = √(9² + 7²) = √130

AC = √130 cm

∴ AM = AC/2 = (√130)/2 cm

Alternatively, AM = √((9/2 cm)² + (7/2 cm)²) = √(32.5) cm

∠TAC = ∠TAM

By trigonometric ratios, we have;

cos (\theta) = \dfrac{Length \ of \ adjacent \ side \ to \ angle }{Length \ of \ hypotenuse\ side \ to \ angle}

\therefore cos (\angle TAM) = cos (\angle TAC) =  \dfrac{\left (\dfrac{\sqrt{130} }{2}   \right )}{12} = \dfrac{\sqrt{130} }{2 \times 12} = \dfrac{\sqrt{130} }{24}

\angle TAC = arccos \left ( \dfrac{\sqrt{130} }{24} \right ) \approx  61.6 ^{\circ} \ to 1 \ decimal \ place

6 0
2 years ago
A circle passes through points A(7,4), B(10,6), C(12,3). Show that AC must be the diameter of the circle.
Artist 52 [7]

so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.

in short, half the distance of AC must be equals to the distance of B to the midpoint of AC, if indeed AC is the diameter.

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{12}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{12+7}{2}~~,~~\cfrac{3+4}{2} \right)\implies \left( \cfrac{19}{2}~~,~~\cfrac{7}{2} \right)=M\impliedby \textit{center of the circle}

now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}}) \\\\\\ BM=\sqrt{\left( \frac{19}{2}-10 \right)^2+\left( \frac{7}{2}-6 \right)^2} \\\\\\ BM=\sqrt{\left( -\frac{1}{2}\right)^2+\left( -\frac{5}{2} \right)^2}\implies \boxed{BM\approx 2.549509756796392}

6 0
3 years ago
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