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3 years ago

Need some help. Its hard.

Mathematics

1 answer:

MArishka [77]3 years ago

7 0

Obvious. All you have to do is divide the equilified factor from the inblified decimal, and incorperate the imberal number from the outside jenquizical factor. so, all you are left with is 2.05 and 1.72. then you have to divide the terms of them, and you are done. Your answer is c

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Answer? to this problem please

Radda [10]

Answer:

Step-by-step explanation:

There is only one vertical asymptote, x = 1.

Only D has vertical asymptote, x = 1.

8 0

2 years ago

the sugar is sold at rupees 5.4 per kg at a loss of 10% at what price relative is sold per kg to earn a profit of 20%

ladessa [460]

Answer:

the price sold per kg to earn a profit of 20% is 7.2 kg

Step-by-step explanation:

The computation of the price sold per kg to earn a profit of 20% is shown below:

But before that the normal price per kg is

= 5.4 per kg × 100 ÷ 90

= 6 per kg

Now for 20% profit, the price per kg is

= 6 × (1 + 0.20)

= 6 + 1.2

= 7.2 kg

hence, the price sold per kg to earn a profit of 20% is 7.2 kg

5 0

3 years ago

Absolute value -1+5 |n|=44 pls show work

dimulka [17.4K]

Answer:

n = -9 or 9

Step-by-step explanation:

-1 + 5|n| = 44

~Add 1 to both sides

5|n| = 45

~Divide 5 to both sides

|n| = 9

~Knowing the absolute value rule, the answer can be positive or negative.

Best of Luck!

4 0

3 years ago

Read 2 more answers

Looking at the top of tower A and base of tower B from points C and D, we find that ∠ACD = 60°, ∠ADC = 75° and ∠ADB = 30°. Let t

katrin2010 [14]

Answer:

$\text{Exact: }AB=25\sqrt{6},\\\text{Rounded: }AB\approx 61.24$

Step-by-step explanation:

We can use the Law of Sines to find segment AD, which happens to be a leg of $\triangle ACD$ and the hypotenuse of $\triangle ADB$ .

The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:

$\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}$

Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is $\angle CAD$ . The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:

$\angle CAD+\angle ACD+\angle CDA=180^{\circ},\\\angle CAD+60^{\circ}+75^{\circ}=180^{\circ},\\\angle CAD=180^{\circ}-75^{\circ}-60^{\circ},\\\angle CAD=45^{\circ}$

Now use this value in the Law of Sines to find AD:

$\frac{AD}{\sin 60^{\circ}}=\frac{100}{\sin 45^{\circ}},\\\\AD=\sin 60^{\circ}\cdot \frac{100}{\sin 45^{\circ}}$

Recall that $\sin 45^{\circ}=\frac{\sqrt{2}}{2}$ and $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ :

$AD=\frac{\frac{\sqrt{3}}{2}\cdot 100}{\frac{\sqrt{2}}{2}},\\\\AD=\frac{50\sqrt{3}}{\frac{\sqrt{2}}{2}},\\\\AD=50\sqrt{3}\cdot \frac{2}{\sqrt{2}},\\\\AD=\frac{100\sqrt{3}}{\sqrt{2}}\cdot\frac{ \sqrt{2}}{\sqrt{2}}=\frac{100\sqrt{6}}{2}={50\sqrt{6}}$

Now that we have the length of AD, we can find the length of AB. The right triangle $\triangle ADB$ is a 30-60-90 triangle. In all 30-60-90 triangles, the side lengths are in the ratio $x:x\sqrt{3}:2x$ , where $x$ is the side opposite to the 30 degree angle and $2x$ is the length of the hypotenuse.

Since AD is the hypotenuse, it must represent $2x$ in this ratio and since AB is the side opposite to the 30 degree angle, it must represent $x$ in this ratio (Derive from basic trig for a right triangle and $\sin 30^{\circ}=\frac{1}{2}$ ).