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lyudmila [28]
2 years ago
15

HELP PLEEEAAAASSSEEEE!!!!

Mathematics
1 answer:
masha68 [24]2 years ago
7 0

Answer:

A) AAS

Step-by-step explanation:

The two pairs of angles are given and they also share a side, meaning that side is congruent to itself, so they are congruent because of Angle Angle Side (AAS)

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AURORKA [14]

Answer:

50.528º

Step-by-step explanation:

tanA = BC/BA

tanA = 17/14

A = 50.528º

3 0
3 years ago
Find the coordinates of point U.
Vesna [10]

Answer:

B(1.4,0.4)

write x axis then y axis

it have 5line between 1 and 2

so 1/5=0.2

one line mean0.2

7 0
3 years ago
Suppose you have $10,000 to invest. Which of the two rates would yield the larger amount in 4 years. 7% compounded quarterly or
GalinKa [24]
K=\$10,000\ \ \ and\ \ \ Sum=K\cdot(1+ \frac{p}{100})^n\\----------------------- \\1)\ \ \ 7\%\ quarterly\\\\ \ \Rightarrow\ \  \frac{1}{4} \cdot7\% =1.75\%\ \ annually\ \ \Rightarrow\ \ p=1.75\\\\ quarterly\ \Rightarrow\ \  4\ times\ annually\ \Rightarrow\ \ 16\ times\ in\ 4\ years\ \Rightarrow\ \ n=16\\\\Sum(7\%)=\$10,000\cdot(1+ \frac{1.75}{100})^{16}=\\ \\.\ \ \ \ \ \ \ \ \ \ \ \ =\$10,000\cdot(1+0,0175)^{16}\approx\$13199.29\\------------------------\\

2)\ \ \ 6.94\%\ daily\\\\ \ \Rightarrow\ \  \frac{1}{365} \cdot6.94\% \approx0.019\%\ \ annually\ \ \Rightarrow\ \ p=0.019\\\\ daily\ \Rightarrow\  365\ times\ annually\ \Rightarrow\ 1420\ times\ in\ 4\ years\ \Rightarrow\ n=1420\\\\Sum(6.94\%)=\$10,000\cdot(1+ \frac{0.019}{100})^{1420}=\\ \\.\ \ \ \ \ \ \ \ \ \ \ \ =\$10,000\cdot(1+0,00019)^{1420}\approx\$13096.69\\---------------------------\\\$13,199.29 > \$13,096.69\\\\Ans.\ the\ larger\ amount\ gives\ the\ compounded\ quarterly.
5 0
3 years ago
In the circle below, suppose m VUX = 152° and mZUVW = 77º. Find the following.
bezimeni [28]

Given:

In the circle, m(\overarc{VUX})=152^\circ and m(\angle MUV)=77^\circ.

To find:

The following measures:

(a) m\angle VUX

(b) m\angle UXW

Solution:

According to the central angle theorem, the central angle is always twice of the subtended angle intercepted on the same same arc.

m(VUX)=2\times m\angle VWX

152^\circ=2\times m\angle VWX

\dfrac{152^\circ}{2}=m\angle VWX

76^\circ=m\angle VWX

In a cyclic quadrilateral, the opposite angles are supplementary angles.

UVWX is a cyclic quadrilateral. So,

m\angle VUX+m\angle VWX=180^\circ          [Opposite angles of a cyclic quadrilateral]

m\angle VUX+76^\circ=180^\circ

m\angle VUX=180^\circ-76^\circ

m\angle VUX=104^\circ

Now,

m\angle UXW+m\angle UVW=180^\circ          [Opposite angles of a cyclic quadrilateral]

m\angle UXW+77^\circ =180^\circ

m\angle UXW=180^\circ-77^\circ

m\angle UXW=103^\circ

Therefore, m\angle VUX=104^\circ  and m\angle UXW=103^\circ .

7 0
3 years ago
Rounding factors to the greatest place results in this
san4es73 [151]
Estimate is the answer
3 0
2 years ago
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