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lapo4ka [179]
3 years ago
14

Can explain that question? please ?& answer

Mathematics
1 answer:
maxonik [38]3 years ago
5 0
If itt says 4×3+6 it is 18 and it means you can either multiply first
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What is the value of 9 cubed?<br><br> 18<br><br> 27<br><br> 81<br><br> 729
exis [7]
Answer is 729

9x9x9 = 729
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3 years ago
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Can someone please help me with this?
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the area of a parallelogram shape land is on the square and length of its two adjacent sides are 25 m and 17 M find its diagonal
Alex787 [66]

Step-by-step explanation:

Draw diagonal AC

The triangle ABC has sides 17 and 25

Say AB is 17, BC is 25

Draw altitude on side BC from A , say h

h = 17 sin B

Area = 25*17 sin B = 408

sin B = 24/25

In ∆ ABC

Cos B = +- 7/25

= 625 + 289 — b^2 / 2*25*17

b^2 = 914 — 14*17 = 676

b = 26

h = 17*24/25 = 408/25 = 16.32

Draw the second diagonal BD

In ∆ BCD, draw altitude from D, say DE =h

BD^2 = h^2 + {(25 + sqrt (289 -h^2) }^2

BD^2 = 16.32^2 + (25 + 4.76)^2

= 885.6576 + 266.3424

BD = √ 1152 = 33.94 m

6 0
3 years ago
Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive
VikaD [51]
Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}&#10;\\\\\\&#10;\textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}

recall that 

\bf sin(\theta)=\cfrac{opposite}{hypotenuse}&#10;\qquad\qquad &#10;cos(\theta)=\cfrac{adjacent}{hypotenuse}&#10;\\\\\\&#10;% tangent&#10;tan(\theta)=\cfrac{opposite}{adjacent}&#10;\qquad \qquad &#10;% cotangent&#10;cot(\theta)=\cfrac{adjacent}{opposite}&#10;\\\\\\&#10;% cosecant&#10;csc(\theta)=\cfrac{hypotenuse}{opposite}&#10;\qquad \qquad &#10;% secant&#10;sec(\theta)=\cfrac{hypotenuse}{adjacent}

therefore, let's just plug that on the remaining ones,

\bf sin(\theta)=\cfrac{-1}{6}&#10;\qquad\qquad &#10;cos(\theta)=\cfrac{\sqrt{35}}{6}&#10;\\\\\\&#10;% tangent&#10;tan(\theta)=\cfrac{-1}{\sqrt{35}}&#10;\qquad \qquad &#10;% cotangent&#10;cot(\theta)=\cfrac{\sqrt{35}}{1}&#10;\\\\\\&#10;sec(\theta)=\cfrac{6}{\sqrt{35}}

now, let's rationalize the denominator on tangent and secant,

\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}&#10;\\\\\\&#10;sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}
3 0
3 years ago
Which of the following is an example of a method that can be used to model 45%? Select all that apply.
Yuki888 [10]

Answer:

B

Step-by-step explanation:

Since there are 100 squares and you shaded 45 that is 45/100 which is equal to 45%

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2 years ago
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