0
What else can I help you with?
Jen has thirteen Q coins and Tom has five Q coins and seven E coins Jennifer has sixty five cents and Tom has three dollars and seventy five cents What is the value of each type of coins show all work?
13Q = 65 divide each side by 13. Q = 5 5Q + 7E = 375 (5*5) + 7E = 375 25 + 7E = 375 Take away 25 from each side 7E = 350 Divide by 7 E = 50
James has 33 coins in his pocket all of them nickels and quarters If he has a total of 2.65 how many quarters does he have?
This problem can be solved by solving the system of equation. Total worth of coins: $2.65 Total number of coins: 33 n= number of nickels q= number of quarters since we know that there are 33 coins total, we can set the equation like this: number of nickels + number of quarters = total number of coins => n+q=33 We also know that the worth of these coins is $2.65. each nickel is worth of $0.05 each quarter is worth of $0.25 therefore we can set the equation: 0.05 x number of nickels + 0.25 x number of quarters = total worth of coins. 0.05n+0.25q=2.65 However, for convienience, we should multiply the equation above by 100 to get rid of decimals. Thus it is 5n+25q=265 you will now have a following set of 2 equations: n+q=33 5n+25q=265 Use the SUBSTITUTION METHOD to solve either n or q for solving n: (replace q with n if you're willing to solve q instead) n+q=33 => n=33-q (since n is equal to 33-q, we can -q -q substitue n in the other equation.) 5(33-q)+25q=265 => 165-5q+25q=265 => 20q=100 => q=5 -165 -165 /20 /20 There are 5 quareters as a result.(or 28 nickels) since you know that q=5 you can substitute q in the first equation. n+(5)=33 => n=28 - 5 -5 therefore, there are 5 quarters and 28 nickels. ELIMINATION METHOD: n x -5 + q x -5 = 33 x -5 => -5n-5q=-165 5n+25q=265 + 5n+25q=265 ------------- 20q=100 => q=5 /20 /20 Or simply we can say: if we have x quarters, we have .25x value of them. So the value of nickels will be 2.65 - .25x. Since we have 33 coins, and x quarters, then the number of nickels will be 33 - x. So the value of all nickels would be also .05(33 - x). Thus, we have:2.65 - .25x = .05(33 - x)2.65 - .25x = 1.65 - .05x2.65 - 1.65 - .25x + .25x = 1.65 - 1.65 - .05x + .25x1 = .20x1/.20 = .20x/.20x = 5 the number of quarters 33 - x= 33 - 5= 28 the number of nickels. Thus, we have 5 quarters and 28 nickels.
What is p plus 3p plus 2q plus 5q?
4p+7q
What is 3p 5q-2p q simplified?
To simplify the expression (3p \cdot 5q - 2p \cdot q), first calculate the products: (3p \cdot 5q = 15pq) and (2p \cdot q = 2pq). Now, subtract the second term from the first: (15pq - 2pq = 13pq). Therefore, the simplified expression is (13pq).
7p plus 5q equals -3 find q when p equals 1?
7+5q = -3 5q = -3-7 5q =-10 q = -2
What is 5q plus 5q?
Oh, isn't that just lovely? When you have 5q plus 5q, you can combine them to make 10q. It's like adding 5 happy little clouds to another 5 happy little clouds to create a beautiful sky full of 10 happy little clouds. Just remember, there are no mistakes in math, only happy little accidents.
What is the answer forSimplify by combining like terms 3q plus 4r-s plus 5q-6r plus 2s?
3q+4r-s+5q-6r+2s
What is 7p plus 5q equals -3 find q when p equals 1?
7p+5q=-3, p=1 7Xp+5q=-3 7+5q=-3 5p=-10 p=-2
How do you factor 4p plus 10q?
2(2p + 5q)
What is -5q-4 plus q equals 12?
q=-4.
What is - 6p - 5q plus 4 - 10p plus q - 3?
To combine the expression (-6p - 5q + 4 - 10p + q - 3), first, group the like terms. For the (p) terms: (-6p - 10p = -16p). For the (q) terms: (-5q + q = -4q). Finally, for the constant terms: (4 - 3 = 1). The simplified expression is (-16p - 4q + 1).
What is the factor of 10q-25q2?
5q(2 - 5q)
Simplify and solve this equation for q 3q plus 5 plus 2q 5 65?
3q + 5 + 2q + 5 = 65 5q + 10 = 65 5q = 55 q = 11 Check it. 33 + 5 + 22 + 5 = 65 It checks.
What is 5q-29 equals 21?
5q-29 equals 21 5q equals 50 q = 5
