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yarga [219]
3 years ago
14

Can you please answer this

Mathematics
2 answers:
Crazy boy [7]3 years ago
5 0
13. 71.5=5.5u divide each side by 5.5 to solve for u
71.5/5.5=13 so u=13

14. 5/6x=5/14 to solve for x you divide both sides by 5/6.  But to divide fractions you multiply the reciprical.  So:
x=5/14*6/5=30/70 or 3/7

15. 44=t/11 multiply both sides by 11 to solve for t
484=t

16. 8.3=b/14 multiply both sides by 14 to solve for b
116.2=b

17. x/3= 15 multiply both sides by 3 to solve for x
x=45

18. 7.8x=23.4 divide both sides by 7.8 to solve for x
x=3
notka56 [123]3 years ago
4 0
13) divide 71.5 by 5.5 to find the answer for u.
14) find the common denominator then divide like in the last one
15) multiply 11 by 44 to find t
16) multiply 14 by 8.3 to find b
17) multiply 3 by 15 to find x
18) divide 23.4 by 7.8 to find x
You might be interested in
Name two properties used to evaluate 7x1-4x1/4
Kitty [74]
7x1=7, then 4x1/4= 4x1=4 divided by 4= 1. Then subtract 7- 1= 6
5 0
3 years ago
What are the four points of intersection between 4x^2 + y^2 - 4y - 32 = 0 and x^2 - y - 7 = 0 ? Solve algebraically
tensa zangetsu [6.8K]
Remember (a²-b²)=(a-b)(a+b)

solve for  a single variable
solve for y in 2nd

add y to both sides
x²-7=y
sub (x²-7) for y in other equaiton

4x²+(x²-7)²-4(x²-7)-32=0
expand
4x²+x⁴-14x²+49-4x²+28-32=0
x⁴-14x²+45=0
factor
(x²-9)(x²-5)=0
(x-3)(x+3)(x-√5)(x+√5)=0
set each to zero

x-3=0
x=3

x+3=0
x=-3

x-√5=0
x=√5

x+√5=0
x=-√5


sub back to find y

(x²-7)=y

for x=3
9-7=2
(3,2)

for x=-3
9-7=2
(-3,2)

for √5
5-7=-2
(√5,-2)

for -√5
5-7=-2
(-√5,-2)


the intersection points are

(3,2)
(-3,2)
(√5,-2)
(-√5,-2)
8 0
4 years ago
How to solve for h? I don't understand it
Vesnalui [34]
H is the altitude of a right triangle, so h²=AD*DC=(20-4)*4=64
h=√64=8
D is the correct answer. 
6 0
3 years ago
Segment AB has length a and is divided by points P and Q into AP , PQ , and QB , such that AP = 2PQ = 2QB. A) Find the distance
Deffense [45]

Answer:

A) \frac{7}{8}a

B) \frac{5}{8}a

Step-by-step explanation:

AB has length a and is divided by points P and Q into AP , PQ , and QB , such that AP = 2PQ = 2QB

A) Therefore, AP = 2QB

QB = AP/2

The midpoint of QB = QB/2 = (AP/2)/2 = AP/4

AP = 2PQ, Therefore PQ = AP/2

Since the length of AB = a

AB = AP + PQ + QB = a

AP + AP/2 + AP/2 = a

AP + AP = a

2AP = a

AP = a/2

The distance between point A and the midpoint of segment QB = AP + PQ + QB/2 = AP + AP/2 + AP/4 = 7/4(AP)

But AP = a/2

Therefore The distance between point A and the midpoint of segment QB =  7/4(a/2)= \frac{7}{8}a

B)

the distance between the midpoints of segments AP and QB = AP/2 + PQ + QB/2 = AP/2 + AP/2 + AP/4 = 5/4(AP)

But AP = a/2

Therefore the distance between the midpoints of segments AP and QB = 5/4(AP) = \frac{5}{4} *\frac{a}{2}=\frac{5}{8}a

7 0
3 years ago
Please help i dont get this at all
Tom [10]

Answer:

\frac{20y^{2} a^{2} }{41bx^{3} }

Step-by-step explanation:

1) Flip the second fraction and change the operation to multiplication:

\frac{5x^{2}y^{3}  }{2a^{5}b^{4}  } *  \frac{8a^{7}b^{3}  }{41x^{5}y  }

2) Cross cancel factors:

\frac{5y^{2}   }{2b  } *  \frac{8a^{2}  }{41x^{3}  }

3) Multiply:

\frac{40y^{2} a^{2} }{82bx^{3} }

4) Simplify:

\frac{20y^{2} a^{2} }{41bx^{3} }

4 0
3 years ago
Read 2 more answers
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