The bisector of an angle of a triangle divides the opposite side into segments that are proportional to the adjacent sides
VT/TK = VY/YK
95.2/168 = 34/YK
YK = (168·34)/95.2 = 60 cm
x = VY + YK = 34 + 60 = 94 cm

7 0

3 years ago

If f (1)= -2, then the point is on the graph of f

Komok [63]

<h3>Answer: (1, -2)</h3>

Explanation:

Writing f(1) = -2 means x = 1 and y = f(x) = -2 pair up together

So we have (x,y) = (1, -2) as the point on the f(x) curve.

6 0

2 years ago

Need a graphing calculator for math class he finds a $60 model on sale for 15% off if he decides to buy the calculator how much

miss Akunina [59]

$60-15%=$51 so $51 is the answer

6 0

2 years ago

Which expression is equivalent to RootIndex 3 StartRoot 32 x Superscript 8 Baseline y Superscript 10 Baseline EndRoot?

evablogger [386]

Answer:

The equivalent expression to the given expression is $\sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}$

Step-by-step explanation:

The given expression is $\sqrt[3]{32x^8y^{10}}$

To find the equivalent expression:

$\sqrt[3]{32x^8y^{10}}=(32x^8y^{10})^{\frac{1}{3}}$

We may write the above expression as below:

$=(32^{\frac{1}{3}})((x^8)^{\frac{1}{3}})((y^{10})^{\frac{1}{3}})$

$=(2)((4)^{\frac{1}{3}})(x^6)\times (x^2)(x^{\frac{2}{3}})(y^3)(y^{\frac{1}{3}})$ (using square root properties)

$=(2\sqrt[3]{4})(x^2\sqrt[3]{x^2})(y^3\sqrt[3]{y})$ (combining the like terms and doing multiplication )

$=2x^2y^3\sqrt[3]{4x^2y}$

Therefore $\sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}$

Therefore the equivalent expression to the given expression is $\sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}$

5 0

2 years ago

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