All categories

2 years ago

amid has gained weight he now weighs 88kg which is 10% higher than his normal weight. what is his normal weight?

Mathematics

1 answer:

dybincka [34]2 years ago

5 0

Answer:

88*0.9= 79.2 (since it's 10% higher, multiply by .9 to find out what 90% of 88 is

you also could've done 88*0.1=8.8 and then subtract 8.8 from 88 and get 79.2

either way his normal weight is 79.2kg

You might be interested in

Can someone please answer this question please answer it correctly and please show work please answer it correctly

liberstina [14]

Answer:

I have attached a photo containing my work and answers. If you have any questions, please let me know!

~Hope this helps!~

7 0

2 years ago

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 $\sin 30^{\circ}=\frac{1}{2}$ ).

Therefore, AB must be exactly half of AD:

$AB=\frac{1}{2}AD,\\AB=\frac{1}{2}\cdot 50\sqrt{6},\\AB=\frac{50\sqrt{6}}{2}=\boxed{25\sqrt{6}}\approx 61.24$

3 0

2 years ago

Read 2 more answers

mr tonneman buys one 2 pound lobster and 1 pound 4 ounces of shrimp. find the ratio the weight of the lobster to the shrimp

Phantasy [73]

2:1over4 this would be the ratio answer

5 0

2 years ago

64.8=6(m+2.9) <br> what is m<br> please help me and explain how did you get your answer

BabaBlast [244]

Answer:

= −11.2

Step-by-step explanation:

64.8=6(m+22)

We move all terms to the left:

64.8-(6(m+22))=0

We calculate terms in parentheses: -(6(m+22)), so:

6(m+22)

We multiply parentheses

6m+132

Back to the equation:

-(6m+132)

We get rid of parentheses

-6m-132+64.8=0

We add all the numbers together, and all the variables

-6m-67.2=0

We move all terms containing m to the left, all other terms to the right

-6m=67.2

4 0

2 years ago

Read 2 more answers

Simplify: (1/2)^4 <br><br> please solve this problem

dsp73

Answer:

-1/16

Step-by-step explanation:

5 0

2 years ago