All categories

2 years ago

Huilan is years 15 younger than Thomas. The sum of their ages is 73 . What is Thomas's age?

Mathematics

2 answers:

UNO [17]2 years ago

7 0

thomas age:44

huilan age: 29

Tomtit [17]2 years ago

7 0

Step-by-step explanation:

h = t + 15

t + t + 15 = 73

2t = 73 - 15

2t = 58

t = 29

You might be interested in

What are the roots of f(x) = x2 – 48? –48 and 48 –24 and 24 Negative 8 StartRoot 3 EndRoot and 8 StartRoot 3 EndRoot Negative 4

Ivanshal [37]

Answer:

$x=4\sqrt{3},\:x=-4\sqrt{3}$ are the roots.

Step-by-step explanation:

$x=4\sqrt{3},\:x=-4\sqrt{3}$

Considering the expression

$x^2\:-\:48$

Solving

$x^2-48=0$

$x^2-48+48=0+48$

$x^2=48$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{48},\:x=-\sqrt{48}$

Solving

$x=\sqrt{48}$

= $\sqrt{2^4\cdot \:3}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$

= $\sqrt{3}\sqrt{2^4}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^m}=a^{\frac{m}{n}}$

$\sqrt{2^4}=2^{\frac{4}{2}}=2^2$

$=2^2\sqrt{3}$

$=4\sqrt{3}$

So,

$x=4\sqrt{3}$

Similarly,

$x=-\sqrt{48}=-4\sqrt{3}$

Therefore, $x=4\sqrt{3},\:x=-4\sqrt{3}$ are the roots.

Keywords: roots, expression

Learn more about roots from brainly.com/question/3731376

#learnwithBrainly

9 0

3 years ago

Read 2 more answers

How many 3-digit numbers can you write using only digits 1 and 7? (Of course, digits can repeat.) How many 3-digit numbers can y

Lana71 [14]

Answer:

111,117,177,777,771,711,717,171

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Nicole can type 42 words in minutes.<br> What is her rate in words per 6 minute?

Aneli [31]

Answer:

252

Step-by-step explanation:

6x42=252

8 0

3 years ago

) For which values of x is f '(x) zero? (Enter your answers as a comma-separated list.) x = (No Response) For which values of x

grigory [225]

Answer:

Check below, please

Step-by-step explanation:

Step-by-step explanation:

1.For which values of x is f '(x) zero? (Enter your answers as a comma-separated list.)

When the derivative of a function is equal to zero, then it occurs when we have either a local minimum or a local maximum point. So for our x-coordinates we can say

$f'(x)=0\: at \:x=2, and\: x=-2$

2. For which values of x is f '(x) positive?

Whenever we have

$f'(x)>0$

then function is increasing. Since if we could start tracing tangent lines over that graph, those tangent lines would point up.

$f'(x)>0 \:at [-4,-2) \:and\:(2, \infty)$

3. For which values of x is f '(x) negative?

On the other hand, every time the function is decreasing its derivative would be negative. The opposite case of the previous explanation. So

$f'(x)$

4.What do these values mean?

$f(x) \:is \:increasing\:when\:f'(x) >0\\\\f(x)\:is\:decreasing\:when f'(x)$

5.(b) For which values of x is f ''(x) zero?

In its inflection points, i.e. when the concavity of the curve changes. Since the function was not provided. There's no way to be precise, but roughly

at x=-4 and x=4

6 0

3 years ago

Simplify each expression by distributing and combined bf like terms (SHOW WORK) please

Dafna1 [17]

Answer:

c-39

1. rearrange terms

2.distribute

3.distribute

4. subtract the numbers

5. combine like terms

6.multiply by 1

8 0

3 years ago