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3 years ago

Answer this please, actually only delany, have a heart she needs the points

Mathematics

2 answers:

MariettaO [177]3 years ago

5 0

the answer is a B:9

Step-by-step explanation:

THE ANSWERS FOR u question is a B:9

natta225 [31]3 years ago

3 0

Answer:

Delaney is answering :)

Step-by-step explanation:

it is either a,b,c,or d

it's gotta be one of those

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‼️ PLEASEEE HELPPP ‼️

4vir4ik [10]

Answer:

take out x+60 and = x+54 cross multiply for starters

3 0

3 years ago

A sheet of construction paper is 9 inches long and 11 inches wide. How many 1- inch squares of paper can cut out one sheet of pa

Serggg [28]

Answer:

Step-by-step explanation: 9*11=99

8 0

3 years ago

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The function p(x) = –8x2 – 64x can be written in vertex form p(x) = a(x – h)2 + k, where a =

nydimaria [60]

Means compplete the square on one side

basically get into form $f(x)=a(x-h)^2+k$

ok, group x terms
$p(x)=(-8x^2-64x)$
factor out linear coefient (-8)
$p(x)=-8(x^2+8x)$
take 1/2 of the linear coefient and square it
8/2=4, 4^2=16
add positive and negaitve of that inside parntheaese
$p(x)=-8(x^2+8x+16-16)$
factor perfect square
$p(x)=-8((x+4)^2-16)$
expand
$p(x)=-8(x+4)^2+128$

in vertex form it is $p(x)=-8(x+4)^2+128$

3 0

4 years ago

4. Use the process outlined in the lesson to approximate the number 2√3. Use the approximation √3 ≈ 1.732 050 8.

jeka57 [31]

Answer:

sequence of five intervals

(1) 2 < $2^{\sqrt{3} }$ < $2^{2}$

(2) $2^{1.7}$ < $2^{\sqrt{3} }$ < $2^{1.8}$

(3) $2^{1.73}$ < $2^{\sqrt{3} }$ < $2^{1.74}$

(4) $2^{1.732}$ < $2^{\sqrt{3} }$ < $2^{1.733}$

(5) $2^{1.7320}$ < $2^{\sqrt{3} }$ < $2^{1.7321}$

Step-by-step explanation:

as per question given data

√3 ≈ 1.732 050 8

to find out

sequence of five intervals

solution

as we have given that √3 value that is here

√3 ≈ 1.732 050 8 ........................1

when we find $2^{\sqrt{3} }$ ................2

put here √3 value in equation number 2

we get $2^{\sqrt{3} }$ that is 3.322

sequence of five intervals

(1) 2 < $2^{\sqrt{3} }$ < $2^{2}$

(2) $2^{1.7}$ < $2^{\sqrt{3} }$ < $2^{1.8}$

(3) $2^{1.73}$ < $2^{\sqrt{3} }$ < $2^{1.74}$

(4) $2^{1.732}$ < $2^{\sqrt{3} }$ < $2^{1.733}$

(5) $2^{1.7320}$ < $2^{\sqrt{3} }$ < $2^{1.7321}$

8 0

3 years ago

Pleasee helppp 15 pointss pleaseee

Leviafan [203]

Let,
The first triangle with sides 7 and 3.5 be ΔABC
The second triangle with sides 3 and x be ΔCDE

ΔABC ≅ ΔCDE (they are congruent) as:
1) They have a common point C
2) ∠ABC = ∠CDE = 90° (given)

Hence we can say,

AC/CE = BC/CD

= 7/x = 3.5/3

Therefore,
x = (7*3)/3.5
= 21/3.5
= 6
Easyyy

8 0

3 years ago

Read 2 more answers