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3 years ago

CAN SOMEONE HELP WITH NUMBER 6&7

Mathematics

1 answer:

mart [117]3 years ago

3 0

Answer:

they are both congruent

Step-by-step explanation:

6 is turned counter clockwise 360 deg

7 is turned counter clockwise 90 deg

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2(t+1)=10 t=?<br><br>please help

bogdanovich [222]

Answer:

Step-by-step explanation:

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3 years ago

Plz help. Giving out brainlist.

tensa zangetsu [6.8K]

Answer:

i think its 2 hours

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3 years ago

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Store A sells a watch for $50 and offers a 5% discount Store B sells the same watch for $60 and offers a 20% discount. From whic

erastova [34]

Answer:

<em>You should buy the watch from Store A because it's cheaper</em>

Step-by-step explanation:

Store A sells a watch for $50 with a 5% discount.

Let's calculate the sale price. The 5% discount is

5/100*$50 = $2.50

The sale price is $50 - $2.50 = $47.50

Now for Store B. The price of the watch was $60 with a 20% discount.

Calculate 20% of $60:

20/100*$60 = $12

The sale price is $60 - $12 = $48

Thus, you should buy the watch from Store A because it's cheaper

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3 years ago

Prove the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus. use the distance formula

riadik2000 [5.3K]

In order to prove this, we have to put the trapezoid to the coordinate system. In the attached photo you can see how it has to be put. The coordinates for the vertices of trapezoid written according to the midpoint principle. By using the distance between two points formula, we can find the coordinates for the vertices of the rhombus.
$D_{x} = \frac{-2b-2a}{2}=-b-a$ and $D_{y}= \frac{2c+0}{2}=c$ . The coordinates of D is $(-b-a, c)$
$E_{x} = \frac{-2b+2b}{2}=0$ and $E_{y}= \frac{2c+2c}{2}=2c$ . The coordinates of E is $(0,2c)$
Since we have the reflection in this graph, the coordinates of F is $(b+a, c)$
And the coordinates of G is (0,0).

Using the distance formula, we can find that
$DE= \sqrt{(-b-a)^{2}+ c^{2}}=\sqrt{(b+a)^{2} + c^{2}}$
$EF= \sqrt{(b+a)^{2} + c^{2} }$
$GF= \sqrt{(b+a )^{2} + c^{2} }$
$DG= \sqrt{(b+a)^{2}+ c^{2}$

Since all the sides are equal this completes our proof. Additionally, we can find the distances of EG and DF in order to show that the diagonals of this rhombus are not equal. So that it is not a square, but rhombus.

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3 years ago

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Someone pls help, its on trig ratios.

Mars2501 [29]

Answer:

$BC=12.8,\\AC=15.1$

Step-by-step explanation:

In right triangles only, the tangent of an angle is equal to its opposite side divided by its adjacent side.

Therefore, we have the following equation:

$\tan 58^{\circ}=\frac{BC}{8}$

Solving, we get:

$BC=8\tan 58^{\circ}\approx \boxed{12.8}$

Also in right triangles only, the cosine of an angle is equal to its adjacent side divided by the hypotenuse of the triangle.

$\cos 58^{\circ}=\frac{8}{AC},\\AC=\frac{8}{\cos 58^{\circ}},\\AC\approx \boxed{15.1}$

We can also use the Pythagorean theorem now that we've found BC. However, if you choose to do so, make sure you don't use a rounded value for BC, as that could cause a notable deviation in your final answer.

Using the Pythagorean theorem to verify our answers:

$8^2+(8\tan 58^{\circ})^2=\left(\frac{8}{\cos 58^{\circ}}\right)^2\:\checkmark$

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3 years ago