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2 years ago

Need help with this geometric question ASAP, Thank you

Mathematics

1 answer:

Law Incorporation [45]2 years ago

7 0

The error would be assuming the altitude is a bisector and divides the sides evenly.

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-7 over 1 3rd written as a fraction is

m_a_m_a [10]

Answer: 22/3 I think

sorry if it’s wrong

7 0

2 years ago

Read 2 more answers

Anyone can help me with my math homework please?

Vedmedyk [2.9K]

Answer:

Step-by-step explanation:

hello,

so we know y in terms of t and x in terms of t and we need to find y in terms of x

$x=21t^2\sqrt{x}=\sqrt{21}*t \ \ as \ \ t>=0 \ \ So\\t=\sqrt{\dfrac{x}{21}}$

and then

$y=f(x)=3\sqrt{\dfrac{x}{21}}+5=\sqrt{\dfrac{9x}{21}}+5=\sqrt{\dfrac{3x}{7}}+5$

hope this helps

4 0

3 years ago

Please help! This assignment is due soon! (In The photo)

eimsori [14]

Answer

Elena must have substracted 1/2x from both sides of the equation.

Lin must have multiplied both sides of the equation by 2

Explanation

The equation given is

$\frac{1}{2}x+3=\frac{7}{2}x+5$

For Elena to have arrived at

$3=\frac{7}{2}x-\frac{1}{2}x+5$

Then Elena must have substracted 1/2x from both sides of the equation.

That is;

$\begin{gathered} \frac{1}{2}x+3=\frac{7}{2}x+5 \\ \text{Substracting }\frac{1}{2}x\text{ from both sides of the equation will give Elena first step} \\ \frac{1}{2}x+3-\frac{1}{2}x=\frac{7}{2}x+5-\frac{1}{2}x \\ 3=\frac{7}{2}x-\frac{1}{2}x+5 \end{gathered}$

For Lin to have arrived at

$x+6=7x+10$

It shows Lin must have multiplied both sides of the equation by 2

That is;

$\begin{gathered} \frac{1}{2}x+3=\frac{7}{2}x+5 \\ \text{Multiply both sides of the equation by the lowest common mutiple } \\ \text{of the denominator which is 2.} \\ \frac{1}{2}x(2)+3(2)=\frac{7}{2}x(2)+5(2) \\ x+6=7x+10 \end{gathered}$

4 0

1 year ago

9. In a rhombus whose side measures 34 and the smaller angle is 42°. Find the length of the larger diagonal, to the nearest tent

Veseljchak [2.6K]

Answer:

63.5

Step-by-step explanation:

The diagonals of a rhombus bisect opposite angles.

The longer diagonal bisects the 42° angle

The larger angle of the rhombus is 180 - 42 = 138

Let x = the length of the longer diagonal

You know two sides and the included angle of a triangle.

Using the law of cosines we get

$x^{2} = 34^{2} + 34^{2} -2(34)(34) cos 138\\ = 1156 + 1156 - 2312 cos 138\\ = 4030.150837...\\x = \sqrt{4030.150837} = 63.48$ $x^{2} = 34^{2} + 34^{2} -2(34)(34)cos 138\\ = 1156 + 1156 - 2312 cos 138\\ = 4030.1508...\\ x = \sqrt{4030.1508} \\ = 63.48$ x^2 = 34^2 + 34^2 - 2(34)(34) cos 138