All categories

3 years ago

Please solve this don’t say you can’t see it if you see it answer please only if you know the answer

Mathematics

1 answer:

Mars2501 [29]3 years ago

8 0

Answer:

5) 3 units

6) 9 units

7) 15.5

8) 67

You might be interested in

Which can be a next step in the construction of an angle with a side on line l that is congruent to ∠ABC?

iren2701 [21]

A) use a straightedge to draw a line thought D and F

6 0

3 years ago

The perimeters of the square and the regular pentagon are equal. A square with side lengths of 3x + 3. A pentagon with side leng

Firlakuza [10]

Answer:

40 hope this helps have a great day

6 0

3 years ago

Read 3 more answers

What is value of zero in 42.02?

Kryger [21]

tenths because 42.02 it would be tenths the first decimal is tenths and second is hundreths

7 0

3 years ago

please help me, Prove a quadrilateral with vertices G(1,-1), H(5,1), I(4,3) and J(0,1) is a rectangle using the parallelogram me

mestny [16]

Answer:

Step-by-step explanation:

We are given the coordinates of a quadrilateral that is G(1,-1), H(5,1), I(4,3) and J(0,1).

Now, before proving that this quadrilateral is a rectangle, we will prove that it is a parallelogram. For this, we will prove that the mid points of the diagonals of the quadrilateral are equal, thus

Join JH and GI such that they form the diagonals of the quadrilateral.Now,

JH= $\sqrt{(5-0)^{2}+(1-1)^{2}}=5$ and

GI= $\sqrt{(4-1)^{2}+(3+1)^{2}}=5$

Now, mid point of JH= $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

= $(\frac{5+0}{2},\frac{1+1}{2})=(\frac{5}{2},1)$

Mid point of GI= $(\frac{5}{2},1)$

Since, mid point point of JH and GI are equal, thus GHIJ is a parallelogram.

Now, to prove that it is a rectangle, it is sufficient to prove that it has a right angle by using the Pythagoras theorem.

Thus, From ΔGIJ, we have

$(GI)^{2}=(IJ)^{2}+(JG)^{2}$ (1)

Now, JI= $\sqrt{(4-0)^{2}+(3-1)^{2}}=\sqrt{20}$ and GJ= $\sqrt{(0-1)^{2}+(1+1)^{2}}=\sqrt{5}$

Substituting these values in (1), we get

$5^{2}=(\sqrt{20})^{2}+(\sqrt{5})^{2} }$

$25=20+5$

$25=25$

Thus, GIJ is a right angles triangle.

Hence, GHIJ is a rectangle.

Also, The diagonals GI= $\sqrt{(4-1)^{2}+(3+1)^{2}}=5$ and HJ= $\sqrt{(0-5)^2+(1-1)^2}=5$ are equal, thus, GHIJ is a rectangle.

6 0

3 years ago

From a boat on the lake, the angle of elevation to the top of a cliff is 26°1'. If the base of the cliff is 205 feet from the bo

Zolol [24]

We use the tangent for this

tan 26 1 = opposite / adjacent = h / 205 where h = height of cliff

h = 205 tan 26 1 = 100 feet

7 0

3 years ago

Read 2 more answers