I am a 2-digit number. The sum of my digits is 16. My ten’s digit is greater than my one’s digit. What number am I?

What is the expression for 9 less than the product of 5 and 7

Helga [31]

(5x7)-9 Thisis because product means multiply "Of" mean sa smaller equation. therefor you need the (...) then 9 less than is minus but it has to go after the product since it's subtraction

3 0

3 years ago

Hi I rlly could use some help with this! I'm not the best at math and I'm rlly struggling with this assignment its a couple ques

snow_lady [41]

1. Let y represent the amount of money earned, and x the number of cars they wash. Then

y = 5x

Why? Each car wash costs $5. So no car washes means no earnings; 1 car wash earns $5; 2 washes earn $10; 3 washes earn $15, and so on, so that x car washes generate $5x.

2. No matter how many cars they end up washing, they will have spent $75 to set up the operation, so you subtract 75 from the equation is part (1).

3. Taking into account the setup cost of $75, the profit they make is given by

y = 5x - 75

4. When x = 0, meaning no cars get washed, the profit is -$75, meaning the team loses $75.

5. Set the profit function equal to 0 and solve for x :

5x - 75 = 0

5x = 75

x = 75/5

x = 15

So the team has to wash at least 15 cars to break even.

5 0

3 years ago

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Could I get help please

aalyn [17]

Answer:

$f(x)=-(x-2)^3 +9$

Step-by-step explanation:

the first - is for reflection

-2 is for translating right (its essentially)

+9 is for translation up

lmk if you want be to explain more

6 0

3 years ago

C=1/21.22.23+1/22.23.24+................+1/200.201.202 . = là dấu nhân

Aneli [31]

It looks like you have to find the value of the sum,

$C = \displaystyle \frac1{21\times22\times23} + \frac1{22\times23\times24} + \cdots + \frac1{200\times201\times202}$

so that the n-th term in the sum is

$\dfrac1{(21+(n-1))\times(21+n)\times(21+(n+1))} = \dfrac1{(n+20)(n+21)(n+22)}$

for 1 ≤ n ≤ 180.

We can then write the sum as

$\displaystyle C = \sum_{n=1}^{180} \frac1{(n+20)(n+21)(n+22)}$

Break up the summand into partial fractions:

$\dfrac1{(n+20)(n+21)(n+22)} = \dfrac a{n+20} + \dfrac b{n+21} + \dfrac c{n+22}$

Combine the fractions into one with a common denominator and set the numerators equal to one another:

$1 = a(n+21)(n+22) + b(n+20)(n+22) + c(n+20)(n+21)$

Expand the right side and collect terms with the same power of n :

$1 = a(n^2+43n+462)+b(n^2+42n+440) + c(n^2+41n + 420) \\\\ 1 = (a+b+c)n^2 + (43a+42b+41c)n + 462a+440b+420c$

Then

a + b + c = 0

43a + 42b + 41c = 0

462a + 440b + 420c = 1

==> a = 1/2, b = -1, c = 1/2

Now our sum is

$\displaystyle C = \sum_{n=1}^{180} \left(\frac1{2(n+20)}-\frac1{n+21}+\frac1{2(n+22)}\right)$

which is a telescoping sum. If we write out the first and last few terms, we have

C = 1/(2×21) - 1/22 + 1/(2×23)

… … + 1/(2×22) - 1/23 + 1/(2×24)

… … + 1/(2×23) - 1/24 + 1/(2×25)

… … + 1/(2×24) - 1/25 + 1/(2×26)

… … + … - … + …

… … + 1/(2×198) - 1/199 + 1/(2×200)

… … + 1/(2×199) - 1/200 + 1/(2×201)

… … + 1/(2×200) - 1/201 + 1/(2×202)

Notice the diagonal pattern of underlined and bolded terms that add up to zero (e.g. 1/(2×23) - 1/23 + 1/(2×23) = 1/23 - 1/23 = 0). So, like a telescope, the sum collapses down to a simple sum of just six terms,

C = 1/(2×21) - 1/22 + 1/(2×22) + 1/(2×201) - 1/201 + 1/(2×202)

which we simplify further to

C = 1/42 - 1/44 - 1/402 + 1/404

C = 1,115/1,042,118 ≈ 0.00106994

4 0

3 years ago

In the diagram, ΔGHJ ≅ ΔSTU.

valina [46]

Answer:

m∠J = 67°

m∠T = 67°

m∠G =46°

ST = 5cm

HJ = 3.7cm

Step-by-step explanation:

Given

ΔGHJ ≅ ΔSTU

GH = 5 cm = GJ
m∠GHJ = 67°

=> m∠J = 67° ( that is the property of isosceles triangle because GH = GJ

=> m∠G = 180° - m∠H - m∠J = 180° - 67° - 67° = 46°

Because ΔGHJ ≅ ΔSTU

=> GH = GJ = ST =SU

=> ST = 5 cm

Because ΔGHJ ≅ ΔSTU and GH = GJ = ST =SU

=> TU = HJ = 3.7 cm

=> m∠T = m∠H = 67°

So:

m∠J = 67°

m∠T = 67°

m∠G =46°

ST = 5cm

HJ = 3.7cm

6 0

3 years ago

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